Question

In the adjoining figure, $\angle B={70}^o,\angle C={50}^o$ and AD is the bisector of $\angle A.$ Prove that AB > AD > CD.

Answer 1

Observe that
$\angle A={180}^o-(\angle B+\angle C)$
$={180}^o-({70}^o+{50}^o)={60}^o$.
Hence, $\angle BAD=\angle DAC={30}^o$.
Consider triangle BAD. We can compute $\angle ADB$:
$\angle ADB={180}^o-({70}^o+{30}^o)={80}^o>\angle ABD$
Hence $AB>AD$

In triangle $ADC$, we have

$\angle DAC={30}^o<{50}^o=\angle ACD.$
Again proposition 2 gives $CD<AD$

Hence, $AB>AD>CD$.