Question

$ABCD$ is a parallelogram. If $AB=2AD$ and $P$ is the mid-point of $CD$; prove that : $\angle APB={90}^{\circ }$

What is the value of $\angle CPB-\angle PBA=$

Answer 1

Since ABCD is a parallelogram and opposite sides of the parallelogram are equal therefore, $AB=DC$ and $AD=BC$.

It is given that $AB=2AD$, that is $AD=\frac{1}{2}AB$

As $AD=BC$ therefore, $AD=BC=\frac{1}{2}AB$..........(eqn 1)

Now if P is the mid point of $CD$ and $AB=DC$, then

$DP=PC=\frac{1}{2}DC=\frac{1}{2}AB$..........(eqn 2)

From equations 1 and 2, $AD=DP$ and we know that angles opposite to equal sides are equal thus, $\angle APD=\angle DAP$..........(eqn 3)

SInce $AB\parallel DC$ and $AP$ is a transversal, therefore,

$\angle APD=\angle PAB$ ( ALternate interior angles)..........(eqn 4)

From equations 3 and 4, we get

$\angle DAP=\angle PAB$

$\Rightarrow$$\angle DAP=\angle PAB=\frac{1}{2}\angle A$..........(eqn 5)

And from equations 1 and 2, $BC=PC$ and we know that angles opposite to equal sides are equal thus, $\angle CPB=\angle PBC$..........(eqn 6)

SInce $AB\parallel DC$ and BP is a transversal, therefore, $\angle CPB=\angle PBA$..........(eqn 7)

From equations 6 and 7, we get

$\angle PBC=\angle PBA$

$\Rightarrow$$\angle PBC=\angle PBA=\frac{1}{2}\angle B$..........(eqn 8)

Now since adjacent sides of a parallelogram are supplementary, therefore,

$\angle A+\angle B={180}^{\circ }$

$\Rightarrow$$\frac{1}{2}\angle A+\frac{1}{2}\angle B={90}^{\circ }$

In triangle APB

$\angle APB+\angle PAB+\angle PBA={180}^{\circ }$

$\Rightarrow$$\angle APB+\frac{1}{2}\angle A+\frac{1}{2}\angle B={180}^{\circ }$

$\Rightarrow$$\angle APB+{90}^{\circ }={180}^{\circ }$

$\Rightarrow$$\angle APB={90}^{\circ }$

Hence proved.