Question

In the adjoining figure, AC > AB and AD is the bisector of ∠A. Show that ∠ADC > ∠ADB.

Answer 1

Given:
$$\begin{gathered} AC>AB \\ \Rightarrow \angle B>\angle C \\ \therefore \angle B+\angle BAD>\angle C+\angle CAD(\because \angle BAD=\angle CAD)...\left(i\right) \end{gathered}$$

In triangles ABD and ADC, the sums of the respective angles will be 180$°$.

$$\begin{gathered} \angle B+\angle BAD+\angle BDA=\angle C+\angle ADC+\angle DAC=180° \\ \Rightarrow \angle BDA=180°-\angle B-\angle BAD \\ \mathrm{and} \\ \Rightarrow \angle ADC=180°-\angle C-\angle CAD \end{gathered}$$

Using (i), we get:

$$\begin{gathered} \angle B+\angle BAD>\angle C+\angle CAD \\ \Rightarrow 180°-\angle B-\angle BAD<180°-\angle C-\angle CAD \\ \Rightarrow \angle BDA<\angle ADC \\ \Rightarrow \angle ADB<\angle ADC \end{gathered}$$

Hence, proved.