Class 10 › Mathematics › Chapter
Triangles
Geometric proofs on similarity criteria, proportionality theorems, area ratios, and right-triangle relations.
1. Congruent Figures & Similar Triangles
Congruent Figures ($\cong$)
Two geometric figures are said to be congruent if they have the same shape and the same size. When placed one over the other, they cover each other completely.
$$\text{Shape: Same} \quad \text{and} \quad \text{Size: Same}$$
Similar Figures ($\sim$)
Two geometric figures are said to be similar if they have the same shape, but not necessarily the same size. All congruent figures are similar, but similar figures need not be congruent.
$$\text{Shape: Same} \quad \text{and} \quad \text{Size: Proportional}$$
Definition of Similar Triangles: Two triangles $\triangle ABC$ and $\triangle DEF$ are said to be similar ($\triangle ABC \sim \triangle DEF$) if and only if:
1. Their corresponding angles are equal: $\angle A = \angle D$, $\angle B = \angle E$, $\angle C = \angle F$.
2. Their corresponding sides are in the same ratio: $$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$$
1. Their corresponding angles are equal: $\angle A = \angle D$, $\angle B = \angle E$, $\angle C = \angle F$.
2. Their corresponding sides are in the same ratio: $$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$$
2. Basic Proportionality Theorem (Thales' Theorem)
Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
In $\triangle ABC$, line $DE \parallel BC$ intersecting $AB$ at $D$ and $AC$ at $E$.
Proof
Given: In $\triangle ABC$, a line parallel to side $BC$ intersects other two sides $AB$ and $AC$ at $D$ and $E$ respectively.
To Prove: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
Construction: Join $BE$ and $CD$. Draw $EN \perp AB$ and $DM \perp AC$.
Proof:
1. Area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Taking base $AD$ and altitude $EN$: $$\text{ar}(\triangle ADE) = \frac{1}{2} \times AD \times EN$$ Taking base $DB$ for $\triangle BDE$ (where $EN$ is the exterior altitude): $$\text{ar}(\triangle BDE) = \frac{1}{2} \times DB \times EN$$ Taking their ratio: $$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{--- (1)}$$ 2. Similarly, taking base $AE$ and altitude $DM$: $$\text{ar}(\triangle ADE) = \frac{1}{2} \times AE \times DM$$ And for $\triangle DEC$ with base $EC$ and altitude $DM$: $$\text{ar}(\triangle DEC) = \frac{1}{2} \times EC \times DM$$ Taking their ratio: $$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--- (2)}$$ 3. Triangles $\triangle BDE$ and $\triangle DEC$ stand on the same base $DE$ and lie between the same parallel lines $DE$ and $BC$. Therefore: $$\text{ar}(\triangle BDE) = \text{ar}(\triangle DEC) \quad \text{--- (3)}$$ 4. From equations (1), (2), and (3), the left-hand sides are equal. Hence: $$\frac{AD}{DB} = \frac{AE}{EC}$$ Hence proved.
1. Area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Taking base $AD$ and altitude $EN$: $$\text{ar}(\triangle ADE) = \frac{1}{2} \times AD \times EN$$ Taking base $DB$ for $\triangle BDE$ (where $EN$ is the exterior altitude): $$\text{ar}(\triangle BDE) = \frac{1}{2} \times DB \times EN$$ Taking their ratio: $$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{--- (1)}$$ 2. Similarly, taking base $AE$ and altitude $DM$: $$\text{ar}(\triangle ADE) = \frac{1}{2} \times AE \times DM$$ And for $\triangle DEC$ with base $EC$ and altitude $DM$: $$\text{ar}(\triangle DEC) = \frac{1}{2} \times EC \times DM$$ Taking their ratio: $$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--- (2)}$$ 3. Triangles $\triangle BDE$ and $\triangle DEC$ stand on the same base $DE$ and lie between the same parallel lines $DE$ and $BC$. Therefore: $$\text{ar}(\triangle BDE) = \text{ar}(\triangle DEC) \quad \text{--- (3)}$$ 4. From equations (1), (2), and (3), the left-hand sides are equal. Hence: $$\frac{AD}{DB} = \frac{AE}{EC}$$ Hence proved.
3. Converse of Basic Proportionality Theorem
Statement: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
Proof by contradiction: assuming $DE \nparallel BC$, let $DE' \parallel BC$.
Proof (By Contradiction)
Given: In $\triangle ABC$, line segment $DE$ intersects $AB$ at $D$ and $AC$ at $E$ such that:
$$\frac{AD}{DB} = \frac{AE}{EC} \quad \text{--- (1)}$$
To Prove: $DE \parallel BC$.
Proof:
1. Suppose $DE$ is not parallel to $BC$. Then there must exist a line through $D$ parallel to $BC$. Let that line be $DE'$, intersecting $AC$ at $E'$.
2. By Basic Proportionality Theorem in $\triangle ABC$ with $DE' \parallel BC$: $$\frac{AD}{DB} = \frac{AE'}{E'C} \quad \text{--- (2)}$$ 3. Comparing equation (1) and equation (2): $$\frac{AE}{EC} = \frac{AE'}{E'C}$$ 4. Adding $1$ to both sides: $$\frac{AE}{EC} + 1 = \frac{AE'}{E'C} + 1 \implies \frac{AE + EC}{EC} = \frac{AE' + E'C}{E'C}$$ $$\frac{AC}{EC} = \frac{AC}{E'C} \implies EC = E'C$$ 5. This is possible only when points $E$ and $E'$ coincide.
6. Hence, our assumption was incorrect, and $DE \parallel BC$.
Hence proved.
1. Suppose $DE$ is not parallel to $BC$. Then there must exist a line through $D$ parallel to $BC$. Let that line be $DE'$, intersecting $AC$ at $E'$.
2. By Basic Proportionality Theorem in $\triangle ABC$ with $DE' \parallel BC$: $$\frac{AD}{DB} = \frac{AE'}{E'C} \quad \text{--- (2)}$$ 3. Comparing equation (1) and equation (2): $$\frac{AE}{EC} = \frac{AE'}{E'C}$$ 4. Adding $1$ to both sides: $$\frac{AE}{EC} + 1 = \frac{AE'}{E'C} + 1 \implies \frac{AE + EC}{EC} = \frac{AE' + E'C}{E'C}$$ $$\frac{AC}{EC} = \frac{AC}{E'C} \implies EC = E'C$$ 5. This is possible only when points $E$ and $E'$ coincide.
6. Hence, our assumption was incorrect, and $DE \parallel BC$.
Hence proved.
4. Internal Angle Bisector Theorem
Statement: The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
Bisector $AD$ of $\angle A$; line $CE \parallel DA$ meets $BA$ produced at $E$.
Proof
Given: In $\triangle ABC$, $AD$ is the internal bisector of $\angle BAC$, meeting $BC$ at $D$.
To Prove: $\dfrac{BD}{DC} = \dfrac{AB}{AC}$.
Construction: Through point $C$, draw a line parallel to $DA$ to meet $BA$ produced at point $E$.
Proof:
1. Since $DA \parallel CE$ and transversal $AC$ intersects them: $$\angle DAC = \angle ACE \quad \text{(Alternate interior angles) --- (1)}$$ 2. Since $DA \parallel CE$ and transversal $BE$ intersects them: $$\angle BAD = \angle AEC \quad \text{(Corresponding angles) --- (2)}$$ 3. Since $AD$ bisects $\angle BAC$, $\angle BAD = \angle DAC$.
4. Therefore, from (1) and (2): $$\angle ACE = \angle AEC$$ 5. In $\triangle ACE$, sides opposite to equal angles are equal: $$AE = AC \quad \text{--- (3)}$$ 6. In $\triangle BCE$, since $DA \parallel CE$, by Basic Proportionality Theorem: $$\frac{BD}{DC} = \frac{BA}{AE}$$ 7. Substituting $AE = AC$ from (3): $$\frac{BD}{DC} = \frac{AB}{AC}$$ Hence proved.
1. Since $DA \parallel CE$ and transversal $AC$ intersects them: $$\angle DAC = \angle ACE \quad \text{(Alternate interior angles) --- (1)}$$ 2. Since $DA \parallel CE$ and transversal $BE$ intersects them: $$\angle BAD = \angle AEC \quad \text{(Corresponding angles) --- (2)}$$ 3. Since $AD$ bisects $\angle BAC$, $\angle BAD = \angle DAC$.
4. Therefore, from (1) and (2): $$\angle ACE = \angle AEC$$ 5. In $\triangle ACE$, sides opposite to equal angles are equal: $$AE = AC \quad \text{--- (3)}$$ 6. In $\triangle BCE$, since $DA \parallel CE$, by Basic Proportionality Theorem: $$\frac{BD}{DC} = \frac{BA}{AE}$$ 7. Substituting $AE = AC$ from (3): $$\frac{BD}{DC} = \frac{AB}{AC}$$ Hence proved.
5. Midpoint Theorem (Proved Using BPT)
Statement: The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.
Proof
Given: In $\triangle ABC$, $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$.
To Prove: $DE \parallel BC$ and $DE = \frac{1}{2} BC$.
Proof:
1. Since $D$ is the midpoint of $AB$, $AD = DB \implies \dfrac{AD}{DB} = 1$.
2. Since $E$ is the midpoint of $AC$, $AE = EC \implies \dfrac{AE}{EC} = 1$.
3. Therefore: $$\frac{AD}{DB} = \frac{AE}{EC}$$ 4. By the Converse of BPT, $DE \parallel BC$.
5. In $\triangle ADE$ and $\triangle ABC$: $$\angle ADE = \angle ABC \quad \text{(Corresponding angles)}, \quad \angle DAE = \angle BAC \quad \text{(Common)}$$ $$\triangle ADE \sim \triangle ABC \quad \text{(AA similarity)}$$ 6. Ratio of corresponding sides: $$\frac{DE}{BC} = \frac{AD}{AB} = \frac{AD}{2AD} = \frac{1}{2} \implies DE = \frac{1}{2} BC$$ Hence proved.
1. Since $D$ is the midpoint of $AB$, $AD = DB \implies \dfrac{AD}{DB} = 1$.
2. Since $E$ is the midpoint of $AC$, $AE = EC \implies \dfrac{AE}{EC} = 1$.
3. Therefore: $$\frac{AD}{DB} = \frac{AE}{EC}$$ 4. By the Converse of BPT, $DE \parallel BC$.
5. In $\triangle ADE$ and $\triangle ABC$: $$\angle ADE = \angle ABC \quad \text{(Corresponding angles)}, \quad \angle DAE = \angle BAC \quad \text{(Common)}$$ $$\triangle ADE \sim \triangle ABC \quad \text{(AA similarity)}$$ 6. Ratio of corresponding sides: $$\frac{DE}{BC} = \frac{AD}{AB} = \frac{AD}{2AD} = \frac{1}{2} \implies DE = \frac{1}{2} BC$$ Hence proved.
6. AAA Similarity Criterion with Proof
Statement: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.
Cut $DP = AB$ and $DQ = AC$ on sides $DE$ and $DF$ of $\triangle DEF$.
Proof
Given: Two triangles $\triangle ABC$ and $\triangle DEF$ such that:
$$\angle A = \angle D, \quad \angle B = \angle E, \quad \angle C = \angle F$$
To Prove: $\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$, and hence $\triangle ABC \sim \triangle DEF$.
Construction: Cut $DP = AB$ on $DE$ and $DQ = AC$ on $DF$. Join $PQ$.
Proof:
1. In $\triangle ABC$ and $\triangle DPQ$: $$AB = DP \quad (\text{By construction}), \quad \angle A = \angle D \quad (\text{Given}), \quad AC = DQ \quad (\text{By construction})$$ By **SAS congruence criterion**: $$\triangle ABC \cong \triangle DPQ$$ 2. Since corresponding parts of congruent triangles (CPCT) are equal: $$\angle B = \angle DPQ$$ 3. But it is given that $\angle B = \angle E$. Therefore: $$\angle DPQ = \angle E$$ 4. Since these are corresponding angles for lines $PQ$ and $EF$ intersected by transversal $DE$: $$PQ \parallel EF$$ 5. By Basic Proportionality Theorem in $\triangle DEF$: $$\frac{DP}{PE} = \frac{DQ}{QF} \implies \frac{PE}{DP} = \frac{QF}{DQ}$$ Adding $1$ to both sides: $$\frac{PE + DP}{DP} = \frac{QF + DQ}{DQ} \implies \frac{DE}{DP} = \frac{DF}{DQ}$$ Taking reciprocals and replacing $DP = AB$ and $DQ = AC$: $$\frac{AB}{DE} = \frac{AC}{DF} \quad \text{--- (1)}$$ 6. Similarly, by marking points on $ED$ and $EF$, we can prove: $$\frac{AB}{DE} = \frac{BC}{EF} \quad \text{--- (2)}$$ 7. From (1) and (2): $$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$$ Since corresponding angles are equal and corresponding sides are proportional, $\triangle ABC \sim \triangle DEF$.
Hence proved.
1. In $\triangle ABC$ and $\triangle DPQ$: $$AB = DP \quad (\text{By construction}), \quad \angle A = \angle D \quad (\text{Given}), \quad AC = DQ \quad (\text{By construction})$$ By **SAS congruence criterion**: $$\triangle ABC \cong \triangle DPQ$$ 2. Since corresponding parts of congruent triangles (CPCT) are equal: $$\angle B = \angle DPQ$$ 3. But it is given that $\angle B = \angle E$. Therefore: $$\angle DPQ = \angle E$$ 4. Since these are corresponding angles for lines $PQ$ and $EF$ intersected by transversal $DE$: $$PQ \parallel EF$$ 5. By Basic Proportionality Theorem in $\triangle DEF$: $$\frac{DP}{PE} = \frac{DQ}{QF} \implies \frac{PE}{DP} = \frac{QF}{DQ}$$ Adding $1$ to both sides: $$\frac{PE + DP}{DP} = \frac{QF + DQ}{DQ} \implies \frac{DE}{DP} = \frac{DF}{DQ}$$ Taking reciprocals and replacing $DP = AB$ and $DQ = AC$: $$\frac{AB}{DE} = \frac{AC}{DF} \quad \text{--- (1)}$$ 6. Similarly, by marking points on $ED$ and $EF$, we can prove: $$\frac{AB}{DE} = \frac{BC}{EF} \quad \text{--- (2)}$$ 7. From (1) and (2): $$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$$ Since corresponding angles are equal and corresponding sides are proportional, $\triangle ABC \sim \triangle DEF$.
Hence proved.
7. AA Similarity Criterion with Proof
Statement: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
Proof
Given: In $\triangle ABC$ and $\triangle DEF$, $\angle A = \angle D$ and $\angle B = \angle E$.
To Prove: $\triangle ABC \sim \triangle DEF$.
Proof:
1. In any triangle, the sum of all three interior angles is $180^\circ$: $$\angle A + \angle B + \angle C = 180^\circ \implies \angle C = 180^\circ - (\angle A + \angle B)$$ $$\angle D + \angle E + \angle F = 180^\circ \implies \angle F = 180^\circ - (\angle D + \angle E)$$ 2. Since $\angle A = \angle D$ and $\angle B = \angle E$: $$\angle C = \angle F$$ 3. Thus, all three corresponding angles of $\triangle ABC$ and $\triangle DEF$ are equal ($\angle A = \angle D$, $\angle B = \angle E$, $\angle C = \angle F$).
4. By the **AAA similarity criterion**, $\triangle ABC \sim \triangle DEF$.
Hence proved.
1. In any triangle, the sum of all three interior angles is $180^\circ$: $$\angle A + \angle B + \angle C = 180^\circ \implies \angle C = 180^\circ - (\angle A + \angle B)$$ $$\angle D + \angle E + \angle F = 180^\circ \implies \angle F = 180^\circ - (\angle D + \angle E)$$ 2. Since $\angle A = \angle D$ and $\angle B = \angle E$: $$\angle C = \angle F$$ 3. Thus, all three corresponding angles of $\triangle ABC$ and $\triangle DEF$ are equal ($\angle A = \angle D$, $\angle B = \angle E$, $\angle C = \angle F$).
4. By the **AAA similarity criterion**, $\triangle ABC \sim \triangle DEF$.
Hence proved.
8. SSS Similarity Criterion with Proof
Statement: If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.
Proof
Given: Two triangles $\triangle ABC$ and $\triangle DEF$ such that:
$$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} < 1 \quad \text{--- (1)}$$
To Prove: $\angle A = \angle D, \angle B = \angle E, \angle C = \angle F$, and hence $\triangle ABC \sim \triangle DEF$.
Construction: Cut $DP = AB$ on $DE$ and $DQ = AC$ on $DF$. Join $PQ$.
Proof:
1. Given $\dfrac{AB}{DE} = \dfrac{AC}{DF}$.
Substituting $AB = DP$ and $AC = DQ$: $$\frac{DP}{DE} = \frac{DQ}{DF} \implies \frac{DE}{DP} = \frac{DF}{DQ}$$ Subtracting $1$ from both sides: $$\frac{DE - DP}{DP} = \frac{DF - DQ}{DQ} \implies \frac{PE}{DP} = \frac{QF}{DQ} \implies \frac{DP}{PE} = \frac{DQ}{QF}$$ 2. By the **Converse of BPT** in $\triangle DEF$: $$PQ \parallel EF$$ 3. Corresponding angles are equal: $$\angle DPQ = \angle E \quad \text{and} \quad \angle DQP = \angle F$$ 4. In $\triangle DPQ$ and $\triangle DEF$, by **AA similarity**: $$\triangle DPQ \sim \triangle DEF \implies \frac{DP}{DE} = \frac{DQ}{DF} = \frac{PQ}{EF}$$ 5. Replacing $DP = AB$: $$\frac{AB}{DE} = \frac{PQ}{EF} \quad \text{--- (2)}$$ 6. But from given relation (1): $$\frac{AB}{DE} = \frac{BC}{EF} \quad \text{--- (3)}$$ Comparing (2) and (3): $$\frac{BC}{EF} = \frac{PQ}{EF} \implies BC = PQ$$ 7. Now comparing $\triangle ABC$ and $\triangle DPQ$: $$AB = DP \quad (\text{Construction}), \quad AC = DQ \quad (\text{Construction}), \quad BC = PQ \quad (\text{Proved})$$ By **SSS congruence criterion**: $$\triangle ABC \cong \triangle DPQ$$ 8. Therefore, corresponding angles are equal: $$\angle A = \angle D, \quad \angle B = \angle DPQ = \angle E, \quad \angle C = \angle DQP = \angle F$$ Since corresponding angles are equal, $\triangle ABC \sim \triangle DEF$.
Hence proved.
1. Given $\dfrac{AB}{DE} = \dfrac{AC}{DF}$.
Substituting $AB = DP$ and $AC = DQ$: $$\frac{DP}{DE} = \frac{DQ}{DF} \implies \frac{DE}{DP} = \frac{DF}{DQ}$$ Subtracting $1$ from both sides: $$\frac{DE - DP}{DP} = \frac{DF - DQ}{DQ} \implies \frac{PE}{DP} = \frac{QF}{DQ} \implies \frac{DP}{PE} = \frac{DQ}{QF}$$ 2. By the **Converse of BPT** in $\triangle DEF$: $$PQ \parallel EF$$ 3. Corresponding angles are equal: $$\angle DPQ = \angle E \quad \text{and} \quad \angle DQP = \angle F$$ 4. In $\triangle DPQ$ and $\triangle DEF$, by **AA similarity**: $$\triangle DPQ \sim \triangle DEF \implies \frac{DP}{DE} = \frac{DQ}{DF} = \frac{PQ}{EF}$$ 5. Replacing $DP = AB$: $$\frac{AB}{DE} = \frac{PQ}{EF} \quad \text{--- (2)}$$ 6. But from given relation (1): $$\frac{AB}{DE} = \frac{BC}{EF} \quad \text{--- (3)}$$ Comparing (2) and (3): $$\frac{BC}{EF} = \frac{PQ}{EF} \implies BC = PQ$$ 7. Now comparing $\triangle ABC$ and $\triangle DPQ$: $$AB = DP \quad (\text{Construction}), \quad AC = DQ \quad (\text{Construction}), \quad BC = PQ \quad (\text{Proved})$$ By **SSS congruence criterion**: $$\triangle ABC \cong \triangle DPQ$$ 8. Therefore, corresponding angles are equal: $$\angle A = \angle D, \quad \angle B = \angle DPQ = \angle E, \quad \angle C = \angle DQP = \angle F$$ Since corresponding angles are equal, $\triangle ABC \sim \triangle DEF$.
Hence proved.
9. SAS Similarity Criterion with Proof
Statement: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar.
Proof
Given: In $\triangle ABC$ and $\triangle DEF$, $\angle A = \angle D$ and:
$$\frac{AB}{DE} = \frac{AC}{DF} < 1 \quad \text{--- (1)}$$
To Prove: $\triangle ABC \sim \triangle DEF$.
Construction: Cut $DP = AB$ on $DE$ and $DQ = AC$ on $DF$. Join $PQ$.
Proof:
1. In $\triangle ABC$ and $\triangle DPQ$: $$AB = DP, \quad \angle A = \angle D, \quad AC = DQ$$ By **SAS congruence criterion**: $$\triangle ABC \cong \triangle DPQ \quad \text{--- (2)}$$ $$\implies \angle B = \angle DPQ \quad \text{and} \quad \angle C = \angle DQP \quad (\text{CPCT})$$ 2. From given relation (1), substitute $AB = DP$ and $AC = DQ$: $$\frac{DP}{DE} = \frac{DQ}{DF} \implies \frac{DP}{PE} = \frac{DQ}{QF}$$ 3. By the **Converse of BPT** in $\triangle DEF$: $$PQ \parallel EF$$ 4. Therefore, corresponding angles are equal: $$\angle DPQ = \angle E \quad \text{and} \quad \angle DQP = \angle F$$ 5. Since $\angle B = \angle DPQ$ and $\angle C = \angle DQP$: $$\angle B = \angle E \quad \text{and} \quad \angle C = \angle F$$ 6. Along with given $\angle A = \angle D$, all three corresponding angles are equal: $$\triangle ABC \sim \triangle DEF \quad (\text{By AAA similarity criterion})$$ Hence proved.
1. In $\triangle ABC$ and $\triangle DPQ$: $$AB = DP, \quad \angle A = \angle D, \quad AC = DQ$$ By **SAS congruence criterion**: $$\triangle ABC \cong \triangle DPQ \quad \text{--- (2)}$$ $$\implies \angle B = \angle DPQ \quad \text{and} \quad \angle C = \angle DQP \quad (\text{CPCT})$$ 2. From given relation (1), substitute $AB = DP$ and $AC = DQ$: $$\frac{DP}{DE} = \frac{DQ}{DF} \implies \frac{DP}{PE} = \frac{DQ}{QF}$$ 3. By the **Converse of BPT** in $\triangle DEF$: $$PQ \parallel EF$$ 4. Therefore, corresponding angles are equal: $$\angle DPQ = \angle E \quad \text{and} \quad \angle DQP = \angle F$$ 5. Since $\angle B = \angle DPQ$ and $\angle C = \angle DQP$: $$\angle B = \angle E \quad \text{and} \quad \angle C = \angle F$$ 6. Along with given $\angle A = \angle D$, all three corresponding angles are equal: $$\triangle ABC \sim \triangle DEF \quad (\text{By AAA similarity criterion})$$ Hence proved.
10. Corresponding Altitudes and Medians in Equiangular Triangles
Result A: Ratio of Sides Equals Ratio of Corresponding Altitudes
Statement: If two triangles are equiangular, then the ratio of their corresponding sides is equal to the ratio of their corresponding altitudes.
Proof:
1. Let $\triangle ABC \sim \triangle DEF$, with altitudes $AL \perp BC$ and $DM \perp EF$.
2. In right triangles $\triangle ALB$ and $\triangle DME$: $$\angle B = \angle E \quad (\text{Since } \triangle ABC \sim \triangle DEF), \quad \angle ALB = \angle DME = 90^\circ$$ 3. By **AA similarity criterion**: $\triangle ALB \sim \triangle DME$.
4. Therefore, corresponding sides are proportional: $$\frac{AB}{DE} = \frac{AL}{DM}$$ Hence proved.
1. Let $\triangle ABC \sim \triangle DEF$, with altitudes $AL \perp BC$ and $DM \perp EF$.
2. In right triangles $\triangle ALB$ and $\triangle DME$: $$\angle B = \angle E \quad (\text{Since } \triangle ABC \sim \triangle DEF), \quad \angle ALB = \angle DME = 90^\circ$$ 3. By **AA similarity criterion**: $\triangle ALB \sim \triangle DME$.
4. Therefore, corresponding sides are proportional: $$\frac{AB}{DE} = \frac{AL}{DM}$$ Hence proved.
Result B: Ratio of Sides Equals Ratio of Corresponding Medians
Statement: If two triangles are equiangular, then the ratio of their corresponding sides is equal to the ratio of their corresponding medians.
Proof:
1. Let $\triangle ABC \sim \triangle DEF$, with medians $AP$ and $DQ$ to sides $BC$ and $EF$ respectively.
2. Since $\triangle ABC \sim \triangle DEF$: $$\frac{AB}{DE} = \frac{BC}{EF} \implies \frac{AB}{DE} = \frac{2BP}{2EQ} = \frac{BP}{EQ}$$ 3. Also, $\angle B = \angle E$.
4. By **SAS similarity criterion**: $\triangle ABP \sim \triangle DEQ$.
5. Therefore: $$\frac{AB}{DE} = \frac{AP}{DQ}$$ Hence proved.
1. Let $\triangle ABC \sim \triangle DEF$, with medians $AP$ and $DQ$ to sides $BC$ and $EF$ respectively.
2. Since $\triangle ABC \sim \triangle DEF$: $$\frac{AB}{DE} = \frac{BC}{EF} \implies \frac{AB}{DE} = \frac{2BP}{2EQ} = \frac{BP}{EQ}$$ 3. Also, $\angle B = \angle E$.
4. By **SAS similarity criterion**: $\triangle ABP \sim \triangle DEQ$.
5. Therefore: $$\frac{AB}{DE} = \frac{AP}{DQ}$$ Hence proved.
11. Ratio of Areas of Two Similar Triangles
Statement: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Similar triangles $\triangle ABC \sim \triangle DEF$ with altitudes $AM$ and $DN$.
Proof
Given: $\triangle ABC \sim \triangle DEF$.
To Prove: $\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \left(\dfrac{AB}{DE}\right)^2 = \left(\dfrac{BC}{EF}\right)^2 = \left(\dfrac{AC}{DF}\right)^2$.
Construction: Draw altitudes $AM \perp BC$ and $DN \perp EF$.
Proof:
1. Area of $\triangle ABC = \frac{1}{2} \times BC \times AM$ and Area of $\triangle DEF = \frac{1}{2} \times EF \times DN$. $$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \frac{\frac{1}{2} \times BC \times AM}{\frac{1}{2} \times EF \times DN} = \frac{BC}{EF} \times \frac{AM}{DN} \quad \text{--- (1)}$$ 2. In $\triangle ABM$ and $\triangle DEN$: $$\angle B = \angle E \quad (\text{Since } \triangle ABC \sim \triangle DEF), \quad \angle AMB = \angle DNE = 90^\circ$$ By AA similarity criterion, $\triangle ABM \sim \triangle DEN$. $$\implies \frac{AM}{DN} = \frac{AB}{DE} \quad \text{--- (2)}$$ 3. Since $\triangle ABC \sim \triangle DEF$: $$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} \quad \text{--- (3)}$$ 4. Substituting (2) and (3) into (1): $$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \frac{BC}{EF} \times \frac{BC}{EF} = \left(\frac{BC}{EF}\right)^2$$ 5. Using equation (3), this directly yields: $$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{AC}{DF}\right)^2$$ Hence proved.
1. Area of $\triangle ABC = \frac{1}{2} \times BC \times AM$ and Area of $\triangle DEF = \frac{1}{2} \times EF \times DN$. $$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \frac{\frac{1}{2} \times BC \times AM}{\frac{1}{2} \times EF \times DN} = \frac{BC}{EF} \times \frac{AM}{DN} \quad \text{--- (1)}$$ 2. In $\triangle ABM$ and $\triangle DEN$: $$\angle B = \angle E \quad (\text{Since } \triangle ABC \sim \triangle DEF), \quad \angle AMB = \angle DNE = 90^\circ$$ By AA similarity criterion, $\triangle ABM \sim \triangle DEN$. $$\implies \frac{AM}{DN} = \frac{AB}{DE} \quad \text{--- (2)}$$ 3. Since $\triangle ABC \sim \triangle DEF$: $$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} \quad \text{--- (3)}$$ 4. Substituting (2) and (3) into (1): $$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \frac{BC}{EF} \times \frac{BC}{EF} = \left(\frac{BC}{EF}\right)^2$$ 5. Using equation (3), this directly yields: $$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{AC}{DF}\right)^2$$ Hence proved.
12. Corollaries on Areas of Similar Triangles
Corollary 1: Equal Areas of Similar Triangles Implies Congruence
Statement: If the areas of two similar triangles are equal, then the triangles are congruent.
Proof:
Given $\triangle ABC \sim \triangle DEF$ and $\text{ar}(\triangle ABC) = \text{ar}(\triangle DEF)$. $$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = 1$$ From the area theorem: $$\left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{AC}{DF}\right)^2 = 1$$ $$\implies AB = DE, \quad BC = EF, \quad AC = DF$$ By the **SSS congruence criterion**, $\triangle ABC \cong \triangle DEF$.
Given $\triangle ABC \sim \triangle DEF$ and $\text{ar}(\triangle ABC) = \text{ar}(\triangle DEF)$. $$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = 1$$ From the area theorem: $$\left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{AC}{DF}\right)^2 = 1$$ $$\implies AB = DE, \quad BC = EF, \quad AC = DF$$ By the **SSS congruence criterion**, $\triangle ABC \cong \triangle DEF$.
Corollary 2: Ratio of Areas Equals Square of Ratio of Altitudes
Since $\dfrac{AB}{DE} = \dfrac{AL}{DM}$ (where $AL$ and $DM$ are corresponding altitudes):
$$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{AL}{DM}\right)^2$$
Corollary 3: Ratio of Areas Equals Square of Ratio of Medians
Since $\dfrac{AB}{DE} = \dfrac{AP}{DQ}$ (where $AP$ and $DQ$ are corresponding medians):
$$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{AP}{DQ}\right)^2$$
13. Pythagoras Theorem & Its Converse
Right triangle $ABC$ with $\angle A = 90^\circ$ and altitude $AD \perp BC$.
Pythagoras Theorem Proof
Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Given: $\triangle ABC$ right-angled at $A$ ($\angle BAC = 90^\circ$).
To Prove: $BC^2 = AB^2 + AC^2$.
Construction: Draw $AD \perp BC$.
Proof:
1. In $\triangle DBA$ and $\triangle ABC$: $$\angle BDA = \angle BAC = 90^\circ, \quad \angle B = \angle B \quad (\text{Common})$$ $$\triangle DBA \sim \triangle ABC \quad (\text{AA similarity})$$ $$\frac{BD}{AB} = \frac{AB}{BC} \implies AB^2 = BC \cdot BD \quad \text{--- (1)}$$ 2. In $\triangle DAC$ and $\triangle ABC$: $$\angle CDA = \angle CAB = 90^\circ, \quad \angle C = \angle C \quad (\text{Common})$$ $$\triangle DAC \sim \triangle ABC \quad (\text{AA similarity})$$ $$\frac{CD}{AC} = \frac{AC}{BC} \implies AC^2 = BC \cdot CD \quad \text{--- (2)}$$ 3. Adding equations (1) and (2): $$AB^2 + AC^2 = BC \cdot BD + BC \cdot CD = BC(BD + CD)$$ Since $BD + CD = BC$: $$AB^2 + AC^2 = BC \cdot BC = BC^2$$ Hence proved.
1. In $\triangle DBA$ and $\triangle ABC$: $$\angle BDA = \angle BAC = 90^\circ, \quad \angle B = \angle B \quad (\text{Common})$$ $$\triangle DBA \sim \triangle ABC \quad (\text{AA similarity})$$ $$\frac{BD}{AB} = \frac{AB}{BC} \implies AB^2 = BC \cdot BD \quad \text{--- (1)}$$ 2. In $\triangle DAC$ and $\triangle ABC$: $$\angle CDA = \angle CAB = 90^\circ, \quad \angle C = \angle C \quad (\text{Common})$$ $$\triangle DAC \sim \triangle ABC \quad (\text{AA similarity})$$ $$\frac{CD}{AC} = \frac{AC}{BC} \implies AC^2 = BC \cdot CD \quad \text{--- (2)}$$ 3. Adding equations (1) and (2): $$AB^2 + AC^2 = BC \cdot BD + BC \cdot CD = BC(BD + CD)$$ Since $BD + CD = BC$: $$AB^2 + AC^2 = BC \cdot BC = BC^2$$ Hence proved.
Converse of Pythagoras Theorem
Statement: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
Summary of Proof: Given $\triangle ABC$ where $BC^2 = AB^2 + AC^2$. Construct a right-angled $\triangle PQR$ with $\angle P = 90^\circ$, $PQ = AB$, and $PR = AC$. By Pythagoras theorem, $QR^2 = PQ^2 + PR^2 = AB^2 + AC^2 = BC^2 \implies QR = BC$. By SSS congruence criterion, $\triangle ABC \cong \triangle PQR$. Therefore, $\angle A = \angle P = 90^\circ$.
14. Result 1: $AB^2 + CD^2 = AC^2 + BD^2$
Statement: In a triangle $ABC$, if $AD \perp BC$, prove that $AB^2 + CD^2 = AC^2 + BD^2$.
Proof
1. In right-angled $\triangle ABD$, by Pythagoras Theorem:
$$AB^2 = AD^2 + BD^2 \implies AD^2 = AB^2 - BD^2 \quad \text{--- (1)}$$
2. In right-angled $\triangle ACD$, by Pythagoras Theorem:
$$AC^2 = AD^2 + CD^2 \implies AD^2 = AC^2 - CD^2 \quad \text{--- (2)}$$
3. Equating both expressions for $AD^2$ from (1) and (2):
$$AB^2 - BD^2 = AC^2 - CD^2$$
$$AB^2 + CD^2 = AC^2 + BD^2$$
Hence proved.
15. Result 2: $AD^2 = BD \cdot CD$ (Altitude to Hypotenuse)
Statement: In $\triangle ABC$, $\angle A = 90^\circ$ and $AD \perp BC$. Prove that $AD^2 = BD \cdot CD$.
Proof
1. In right $\triangle ABC$, $\angle BAC = 90^\circ \implies \angle B + \angle C = 90^\circ \implies \angle C = 90^\circ - \angle B$.
2. In right $\triangle ABD$, $\angle BDA = 90^\circ \implies \angle BAD = 90^\circ - \angle B$.
3. Therefore:
$$\angle BAD = \angle C$$
4. Now, in $\triangle DBA$ and $\triangle DAC$:
$$\angle BDA = \angle ADC = 90^\circ$$
$$\angle BAD = \angle ACD$$
By **AA similarity criterion**:
$$\triangle DBA \sim \triangle DAC$$
5. Taking corresponding side ratios:
$$\frac{BD}{AD} = \frac{AD}{CD} \implies AD^2 = BD \cdot CD$$
Hence proved.
16. Result 3: Converse of Right Altitude Result
Statement: In $\triangle ABC$, $AD \perp BC$ such that $AD^2 = BD \cdot CD$. Prove that $\angle A = 90^\circ$.
Proof
1. Given $AD^2 = BD \cdot CD \implies \dfrac{AD}{CD} = \dfrac{BD}{AD}$.
2. In $\triangle DBA$ and $\triangle DAC$:
$$\frac{BD}{AD} = \frac{AD}{CD}, \quad \text{and} \quad \angle BDA = \angle ADC = 90^\circ$$
By **SAS similarity criterion**:
$$\triangle DBA \sim \triangle DAC$$
3. Since corresponding angles are equal:
$$\angle BAD = \angle C \quad \text{and} \quad \angle B = \angle CAD$$
4. Sum of angles in $\triangle ABC$:
$$\angle B + \angle C + \angle A = 180^\circ$$
$$(\angle CAD) + (\angle BAD) + \angle A = 180^\circ$$
Since $\angle CAD + \angle BAD = \angle A$:
$$\angle A + \angle A = 180^\circ \implies 2\angle A = 180^\circ \implies \angle A = 90^\circ$$
Hence proved.
17. Result 4: Acute Angle Theorem ($\angle B < 90^\circ$)
Statement: In $\triangle ABC$, $\angle B$ is acute ($\angle B < 90^\circ$) and $AD \perp BC$. Prove that: $$AC^2 = AB^2 + BC^2 - 2BC \cdot BD$$
Acute-angled triangle $ABC$ with $AD \perp BC$ ($D$ lies on segment $BC$).
Proof
1. In right-angled $\triangle ADC$, by Pythagoras Theorem:
$$AC^2 = AD^2 + CD^2 \quad \text{--- (1)}$$
2. Notice that $CD = BC - BD$. Substituting into (1):
$$AC^2 = AD^2 + (BC - BD)^2$$
$$AC^2 = AD^2 + BC^2 + BD^2 - 2BC \cdot BD$$
$$AC^2 = (AD^2 + BD^2) + BC^2 - 2BC \cdot BD \quad \text{--- (2)}$$
3. In right-angled $\triangle ABD$, by Pythagoras Theorem:
$$AD^2 + BD^2 = AB^2$$
4. Substituting $AD^2 + BD^2 = AB^2$ into equation (2):
$$AC^2 = AB^2 + BC^2 - 2BC \cdot BD$$
Hence proved.
18. Result 5: Obtuse Angle Theorem ($\angle B > 90^\circ$)
Statement: In $\triangle ABC$, $\angle B$ is obtuse ($\angle B > 90^\circ$) and $AD \perp BC$ produced. Prove that: $$AC^2 = AB^2 + BC^2 + 2BC \cdot BD$$
Obtuse-angled triangle $ABC$ with $AD \perp BC$ produced at $D$.
Proof
1. In right-angled $\triangle ADC$, by Pythagoras Theorem:
$$AC^2 = AD^2 + CD^2 \quad \text{--- (1)}$$
2. Notice that $CD = BC + BD$. Substituting into (1):
$$AC^2 = AD^2 + (BC + BD)^2$$
$$AC^2 = AD^2 + BC^2 + BD^2 + 2BC \cdot BD$$
$$AC^2 = (AD^2 + BD^2) + BC^2 + 2BC \cdot BD \quad \text{--- (2)}$$
3. In right-angled $\triangle ADB$, by Pythagoras Theorem:
$$AD^2 + BD^2 = AB^2$$
4. Substituting $AD^2 + BD^2 = AB^2$ into equation (2):
$$AC^2 = AB^2 + BC^2 + 2BC \cdot BD$$
Hence proved.
19. Result 6: Apollonius' Theorem (Median Relation)
Statement: In $\triangle ABC$, if $AD$ is a median, prove that: $$AB^2 + AC^2 = 2(AD^2 + BD^2)$$
Median $AD$ with auxiliary perpendicular $AL \perp BC$.
Proof
Construction: Draw altitude $AL \perp BC$.
1. Median $AD$ makes one acute angle and one obtuse angle with base $BC$ (unless $AD \perp BC$, which is straightforward).
Let $\angle ADB$ be obtuse ($\angle ADB > 90^\circ$), so $\angle ADC$ is acute ($\angle ADC < 90^\circ$).
Let $\angle ADB$ be obtuse ($\angle ADB > 90^\circ$), so $\angle ADC$ is acute ($\angle ADC < 90^\circ$).
2. In $\triangle ABD$, with obtuse angle $\angle ADB$, by the Obtuse Angle Theorem (Result 5):
$$AB^2 = AD^2 + BD^2 + 2BD \cdot DL \quad \text{--- (1)}$$
3. In $\triangle ACD$, with acute angle $\angle ADC$, by the Acute Angle Theorem (Result 4):
$$AC^2 = AD^2 + CD^2 - 2CD \cdot DL \quad \text{--- (2)}$$
4. Since $AD$ is a median, $D$ is the midpoint of $BC$, so $BD = CD$. Substituting $CD = BD$ into (2):
$$AC^2 = AD^2 + BD^2 - 2BD \cdot DL \quad \text{--- (3)}$$
5. Adding equations (1) and (3):
$$AB^2 + AC^2 = (AD^2 + BD^2 + 2BD \cdot DL) + (AD^2 + BD^2 - 2BD \cdot DL)$$
$$AB^2 + AC^2 = 2AD^2 + 2BD^2 = 2(AD^2 + BD^2)$$
Hence proved.