Name: Areas of Similar Triangles
Uploaded: 2019-07-19
Duration: 6 min 54 s
Description: The corresponding sides of similar triangles are proportional. Are the ratios of areas of similar triangles the same as that of their sides? Watch this video...

Question

Prove that ''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''.

Accepted Answer

This theorem can be proved by taking two triangles $A B C$ and $D E F$ such that $∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F$ .

Cut $D P = A B, D Q = A C$ and join $P Q$ .

In $△ A B C$ and $△ D P Q$ ,

$A B = D P$ (by construction)

$A C = D Q$ (by construction)

$∠ A = ∠ D$ (given)

$\Rightarrow △ A B C ≅ △ D P Q$ (By SAS criterion)

$\Rightarrow ∠ B = ∠ P$ (By CPCT)

But $∠ B = ∠ E$ (given)

$\Rightarrow ∠ E = ∠ P$

But $∠ P$ and $∠ E$ are corresponding angles, therefore, $P Q ∥ E F$ .

Therefore, $\frac{D P}{P E} = \frac{D Q}{Q F}$ i.e. $\frac{P E}{D P} = \frac{Q F}{D Q}$

Adding $1$ to both sides gives,

$\frac{P E}{D P} + 1 = \frac{Q F}{D Q} + 1$

$\Rightarrow \frac{P E + D P}{D P} = \frac{Q F + D Q}{D Q}$

$\Rightarrow \frac{D E}{D P} = \frac{D F}{D Q}$

$\Rightarrow \frac{D E}{A B} = \frac{D F}{A C}$

$\Rightarrow \frac{A B}{D E} = \frac{A C}{D F}$

Similarly, $\frac{A B}{D E} = \frac{B C}{E F}$

Therefore, $\frac{A B}{D E} = \frac{A C}{D F} = \frac{B C}{E F}$

Since all $3$ sides are in proportion.

$∴ △ A B C ~ △ D E F$ .

Hence, proved.

Prove that ''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''.

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