Question

In $\Delta$ABC, side AB is produced to D such that BD$=$BC. If $\angle A={70}^o$ and $\angle B={60}^o$, prove that AD$>$CD.

Answer 1

$\angle ACB={50}^{\circ }$ Sum of angles $={180}^{\circ }$

Let $\angle BCD=x,\angle BDC=y$

now, $x+y={60}^{\circ }$ [exterior angle property]

Also, $70+50+x+y={180}^{\circ }$ Sum of angles $={180}^{\circ }$ for triangle ACD

Given $BD=BC$ $\Rightarrow x=y={30}^{\circ }$ Two sides of same length hence same angle.

$\angle ACD={50}^{\circ }+{30}^{\circ }={80}^{\circ }$

Now, in $\Delta ACD,CD$ is the side opposite to $\angle A$ & $AD$ is side opposite to $\angle C$.

$\because \angle C>\angle A$

$\therefore AD>CD[\because$ side opposite to greater angle is longer $]$