Question

ABCD is a parallelogram and $\angle DAB={60}^0.$, If the bisectors AP and BP of the angles A and B respective meet at P on CD. Prove that P is the mid-point of CD.

Answer 1

$ABCD$ is a parallelogram, in which $\angle A={60}^o.$

$\Rightarrow$ $\angle A+\angle B={180}^o$ [ Sum of adjacent angles of parallelogram are supplementary ]

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

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

$\Rightarrow$ $\angle D=\angle B={120}^o$ [ Opposite angles of parallelogram are equal ]

$\Rightarrow$ $\angle A=\angle C={60}^o$ [ Opposite angles of parallelogram are equal ]

$AP$ bisects $\angle A$

$\therefore$ $\angle DAP=\angle PAB={30}^o.$

$BP$ bisects $\angle B$

$\therefore$ $\angle CBP=\angle PBA={60}^o$

In $△PAB,$

$\Rightarrow$ $\angle PAB+\angle PBA+\angle APB={180}^o.$

$\Rightarrow$ ${30}^o+{60}^o+\angle APB={180}^o$

$\Rightarrow$ $\angle APB={90}^o.$

In $△PBC,$

$\Rightarrow$ $\angle PBC+\angle PCB+\angle BPC={180}^o.$

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

$\Rightarrow$ $\angle BPC={60}^o$

In $△ADP,$

$\Rightarrow$ $\angle PAD+\angle ADP+\angle APD={180}^o.$

$\Rightarrow$ ${30}^o+{120}^o+\angle APD={180}^o.$

$\Rightarrow$ $\angle APD={30}^o.$

In $△PBC,$

$\Rightarrow$ $\angle BPC=\angle CBP={60}^o.$ [ linear angles are supplementary ]

$\Rightarrow$ $BC=PC$ [ Sides opposite to equal angles of a triangle are equal ] ----- ( 1 )

In $△ADP,$

$\Rightarrow$ $\angle APD=\angle DAP={30}^o.$

$\Rightarrow$ $AD=DP$ [ Sides opposite to equal angles of triangle are equal ]

But $AD=BC$ [ Opposite sides of parallelogram are equal ]

$\Rightarrow$ So, $BC=DP$ ----- ( 2 )

From ( 1 ) and ( 2 ), we get

$\Rightarrow$ $DP=PC$

$\Rightarrow$ $P$ is the mid-point of $CD$