Question

In the figure alongside,
AB = AC
$\angle A=48$ and
$\angle ACD={18}^{\circ }$.
hence, AB = CD.

• A. True
• B. False

Answer 1

The correct option is B False

Given, AB = AC
$\angle ABC=\angle ACB=x$ (Isosceles triangle property)
In $△ABC$
$\angle ABC+\angle ACB+\angle ABC=180$ (Sum of angles of triangle)
$x+x+48=180$
$2x=132$
$x={66}^{\circ }$
$\angle BCA=\angle BCD+\angle DCB=66$
$\angle BCD=66-18$
$\angle BCD={48}^{\circ }$
Now, In $△BCD$
Sum of angles of triangle = 180
$\angle BCD+\angle BDC+\angle DBC=180$
$48+\angle BDC+66=180$
$\angle BDC={66}^{\circ }$
Since, $\angle BDC=\angle DBC$
hence, $BD=CD$ (Isosceles triangle property)
Thus, $△BDC$ is an isosceles triangle.