Question

In the given figure, ABCD is a rectangle. Diagonals AC and BD intersect each other at P. If $\angle$APD$={52}^o$, find $\angle$ACB and $\angle$ABD.

Answer 1

Since $AC=BD$ and $P$ bisects them

So, $PD=PA$

So, $\angle PAD=\angle PDA=x$ (Let)

In $△PDA$

$\angle APD+\angle PDA+\angle DPA={180}^{\circ }$

So, $\angle PAD=\angle PDA={64}^{\circ }$

Similarly Since $\angle DPA=\angle CPB$ (vertically opposite angles)

$\angle PCB=\angle PBC={64}^{\circ }$

So, $\angle ACB={64}^{\circ }$

Now, $\angle B={90}^{\circ }$ (angle of rectangle)

$\angle CBD+\angle PBA={90}^{\circ }{64}^{\circ }+\angle PBA={90}^{\circ }\angle PBA={26}^{\circ }$

So, $\angle ABD={26}^{\circ }$