In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X.

Given: In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X.

In $\Delta A B X$,

$\because \angle B A X>\angle A B X$

$\therefore B X>A X \quad \ldots(\mathrm{i})$

Similarly, in $\Delta A C X$,

$\because \angle A C X>\angle X A C$

$\therefore A X>C X \quad \ldots$ (ii)

From (i) and $(i i)$, we get

$B X>A X>C X$