Question

The diagonals $AC$ and $BD$ of a rectangle $ABCD$ intersects each other at $P$. If $\angle ABD={50}^o$, find $\angle DPC$.

Answer 1

Given, $ABCD$ is a rectangle
We know that the diagonals of rectangle are same and bisect each other
So, we have
$AP=BP$
$\angle PAB=\angle PBA$ (Equal sides have equal opposite angles)
$\angle PAB={50}^o$ (Since, given $\angle PBA={50}^o$
Now, in $△APB$
$\angle APB+\angle ABP+\angle BAP={180}^o$
$\angle APB+{50}^o+{50}^o={180}^o$
$\angle APB={180}^o-{100}^o$
$\angle APB={80}^o$
Then,
$\angle DPC=\angle APB$ (Vertically opposite angles)
Hence,
$\angle DPC={80}^o$