Question

In figure, ABCD is a parallelogram in which $\angle$A $={60}^o$. If the bisectors of $\angle$A and $\angle$B meet at P, prove that AD$=$DP, PC$=$BC and DC$=2$AD.

Answer 1

$AP$ bisects $\angle A$

Then, $\angle DAP=\angle PAB={30}^o$ ------ ( 1 )

We know that in parallelogram adjacent angles are supplementary

$\therefore$ $\angle A+\angle B={180}^o$

$\Rightarrow$ ${60}^o+\angle B={180}^o$

$\therefore$ $\angle B={120}^o.$

$BP$ bisects $\angle B$

Then, $\angle PAB=\angle PBC={60}^o$ ---- ( 2 )

$\Rightarrow$ $\angle PAB=\angle APD={30}^o$ [ Alternate angles ] ---- ( 3 )

$\therefore$ $\angle DAP=\angle APD={30}^o$ [ From ( 1 ) and ( 3 ) ]

$\therefore$ $AD=DP$ [ Since base angles are equal ]

Similarly, $\angle PBA=\angle BPC={60}^o$ [ Alternate angles ] --- ( 4 )

$\Rightarrow$ $\angle PBC=\angle BPC={60}^o$ [ From ( 2 ) and ( 4 ) ]

$\therefore$ $PC=BC$ [ Since base angles are equal

$\Rightarrow$ $DC=DP+PC$

$\Rightarrow$ $DC=AD+BC$ [ Since, $DP=AD,PC=BC$ ]

$\Rightarrow$ $DC=AD+AD$ [ Opposite sides of parallelogram are equal ]

$\therefore$ $DC=2AD$