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$>$AC.

Answer 1

$\angle A={70}^{\circ },\angle CBA={60}^{\circ },\angle CBD={120}^{\circ }$

$\angle ABC={50}^{\circ },\angle BCD=\angle BDC={30}^{\circ }$ [isoscles triangle]

Now, In ${△}^{le},ADC$

$\angle A={70}^{\circ },\angle B={30}^{\circ },\angle C={80}^{\circ }$

side $AD$ is opposite to $\angle C$ & $AC$ is opposite to $\angle D$

$\because \angle D<\angle C\Rightarrow AC<ADorAD>AC$

[$\because$ side opposite to greater $\angle$ is longer]