Question

In triangleABC, side AB is produced to D so that BD = BC. If angle $B={60}^{\circ }$ and angle $A={70}^{\circ }$ prove that
1.AD>CD
2.AD>AC.

Answer 1

$\angle C={80}^{\circ }$
$\angle A={70}^{\circ }$ [In $\Delta$ ACD]
since $\angle C>\angle A$
(opposite) Corresponding sides i.e.,
AD is greater than CD.
[Since angle opposite to larger side is greater, and side opposite to large angle is larger].
$\Rightarrow$ AD > CD Proved.
$\angle C={80}^{\circ };\angle D={30}^{\circ }$ [In $\Delta$ ACD]
opposite corresponding sides, $AD>AC[\because \angle C>\angle A]$
$\therefore AD>AC$ Proved.