Question

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.

Answer 1

Given: Here $△ABC$ and $△DEF$ are such that

$\angle BAC=\angle EDF$

$\frac{AB}{AC}=\frac{DE}{DF}$

To prove that $△ABC\sim △DEF$

Construction- Draw $D{B}^{\prime }$ equal to $AB$ and $D{C}^{\prime }$ equal to $AC$ in $△DEF$ and join $B^{\prime }{C}^{\prime }$.

Proof - In $△ABC$ and $△D{B}^{\prime }{C}^{\prime }$

$AB=B{D}^{\prime }$ By construction

$AC=D{C}^{\prime }$ by construction

$\angle A=\angle D$ Given

$△ABC\cong △D{B}^{\prime }{C}^{\prime }$ ...(S.A.S test of Congruence)

$\angle B=\angle D{B}^{\prime }{C}^{\prime }$ ...C.A.C.T

$\angle C=\angle D{C}^{\prime }{B}^{\prime }$ ....C.A.C.T

$\therefore △ABC\sim △D{B}^{\prime }{C}^{\prime }$ ....(A.A.A test of similarity)

$\frac{AB}{D{B}^{\prime }}=\frac{AC}{D{C}^{\prime }}=\frac{BC}{B^{\prime }{C}^{\prime }}$ ....(C.S.S.T)

But $\frac{AB}{DE}=\frac{AC}{DF}$ ...Given

Or $\frac{D{B}^{\prime }}{DE}=\frac{D{C}^{\prime }}{DF}$

$\{AB=D{B}^{\prime },AC=D{C}^{\prime }\}$

$\Rightarrow B^{\prime }{C}^{\prime }\parallel EF$.
(Side divides the two side in the same ratio then it is parallel to third side).

$\therefore \angle D{B}^{\prime }{C}^{\prime }=\angle E=\angle B$

$\angle D{C}^{\prime }{B}^{\prime }=\angle F=\angle C$

$\therefore △ABC\sim △DEF$ (S.A.S.test of similarity)