Question

In a triangle ABC, if AB = AC and AB is produced to D such that BD = BC, find $\angle ACD:\angle ADC.$

Answer 1

In $\Delta ABC,AB=AC$

AB is produced to D such that BD = BC DC are joined

In $\Delta ABC,AB=AC$
$\therefore \angle ABC=\angle ACB$
In $\Delta BCD,BD=BC$
$\therefore \angle BDC=\angle BCD$
and Ext.$\angle ABC=\angle BDC+\angle BCD$
$=2\angle BDC$ $(\because \angle BDC=\angle BCD)$
$\Rightarrow \angle ACB=2\angle BCD(\because \angle ABC=\angle ACB)$
$\Rightarrow \angle ACB+\angle BDC=2\angle BDC+\angle BDC$
$\Rightarrow \angle ACB+\angle BCD=3\angle BDC$
$\because \angle BDC=\angle BCD)$
$\Rightarrow \angle ACD=3\angle BDC$
$\therefore \frac{\angle ACD}{\angle BDC}=\frac{3}{1}$
$\therefore$ Ratio between $\angle ACD$ and $\angle BDC=3:1$