Question

Construct an equilateral triangle $\Delta ABC$. Construct an angle bisector of $\angle A$.
Let it meet side $BC$ at $D$. Find measures of $BD$, and $DC$. Then:

• A. $BD>CD$
• B. $BD<CD$
• C. $BD=CD$
• D. None

Answer 1

Answer: (C)

$BD=CD$

In $△ABD\&△ACD$,

AB = AC,

$\angle BAD=\angle CAD$,

AD is common side.

So, $△ABD\cong △ACD$

Hence, BD = CD.

So, option C is correct.