Question

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Answer 1

Its a parallelogram

<C+<D=<A+<C

Also

AB||CD

`\\ \rm\Rrightarrow <A+<B=180`

`\\ \rm\Rrightarrow (1)/(2)<A+(1)/(2)<B=(1)/(2)180`

`\\ \rm\Rrightarrow <PAB+<ABP=90`

Using angle sum property

`\\ \rm\Rrightarrow <APB=180-90=90`

Hence proved