In triangle $A B C$ , the bisector of angle $B A C$ meets opposite side $B C$ at point $D$ . If $B D = C D$ , prove that $Δ A B C$ is isosceles.

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Solution

Given $A D$ is angle bisector and $B D = C D .$

Construction:
Draw a line from $C$ parallel to $A B$ with length $A C .$

Now, in $△ A B D$ and $△ F C D,$
$∠ A B D = ∠ F C D$ [Corresponding Angles]
$∠ B A D = ∠ C F D$ [Corresponding Angles]
So, by AAA criteria of similarity,
$△ A B D \sim △ F C D$
$\Rightarrow \frac{A B}{B D} = \frac{C F}{C D}$

But given $B D = C D$ and we took $C F = A C$
$\Rightarrow A B = A C$

$∴ △ A B C$ is isosceles.

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In triangle ABC, the bisector of angle BAC meets opposite side BC at point D. If BD=CD, prove that ΔABC is isosceles.

In triangle $A B C$ , the bisector of angle $B A C$ meets opposite side $B C$ at point $D$ . If $B D = C D$ , prove that $Δ A B C$ is isosceles.