Question

In parallelogram $ABCD$, the bisector of angle $A$ meets $CB$ in $P$ and the bisector of angle $B$ meets $DA$ in $Q$ and $AB=2AD$.Then $\angle ABP=xCBP$. Find the value of $x$.

Answer 1

Let the $\angle DAB=2\alpha$

$AP$ bisects the angle.

So, $\angle DAP=\angle PAB=\alpha$

As $AD\parallel BC,\angle ABC=180-2\alpha$

$BQ$ bisects $\angle ABC$, So $\angle ABQ=\angle PBQ=90-\alpha$

Let the point of intersection of $BQ$ and $AP$ be $O$

Consider $△AQB$,

$\angle QAB+\angle ABQ+\angle AQB=180°$

$⟹2\alpha +90-a+\angle AQB=180°$

$⟹\angle AQB=90-\alpha$

Similarly, $\angle APB=\alpha$

So, $△QAB$ and $△PAB$ isosceles triangles with $AQ=AB$ and $PB=AB$

$\therefore AQ=PB$

$\angle ABP=CBP$