Question

In a parallelogram ABCD, the bisectors of $\angle A$ also bisect BC at x, find AB : AD.

Answer 1

In || gm ABCD,

Bisectors of $\angle A$ meets BC at X

and BX = XC

Draw XY || gm AB meeting AD at Y

$\because diagonalAXbisects\angle A$

$\therefore$ ABXY is a rhombus

$\therefore AB=BX=AY$

Now $\frac{AB}{AD}=\frac{AB}{BC}=\frac{AB}{2BX}=\frac{BX}{2BX}=\frac{1}{2}\therefore AB:AD=1:2$