Question

In a triangle $ABC$, $AB=AC$ and $\angle A={36}^{\circ }$. If the internal bisector of $\angle C$ meets $AB$ at point $D$, then

• A. AD = BC
• B. AD = AC
• C. AD = AB
• D. AB = AC = BC

Answer 1

The correct option is A AD = BC

Given: $\angle A={36}^{\circ }$ and $AB=AC$
Since, $AB=AC$
$\angle B=\angle C=x$ (Isosceles triangle property)
In $△ABC$
$\angle A+\angle B+\angle C=180$
$36+x+x=180$
$x={72}^{\circ }$
$\angle B=\angle C={72}^{\circ }$
Since, $CD$ bisects $\angle C$
$\angle BCD=\angle ACD=\frac{1}{2}\angle C={36}^{\circ }$
Now, In $△BDC$,
$\angle B+\angle BCD+\angle BDC=180$ (Angle sum property)
$72+36+\angle BDC=180$
$\angle BDC={72}^{\circ }$
Thus, $\angle BDC=\angle B={72}^{\circ }$
Hence, $BC=AD$ (Isosceles triangle property)