Question

In $△ABC$, $AB=AC$ 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, $BC=CD$ ...(Isosceles triangle property) (1)
In$△ADC$
$AD=CD$( isosceles triangle property) (2)
From (1) and (2)
$AD=BC$