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In a parallelogram ABCD, the bisectors o...

In a parallelogram ABCD, the bisectors of `angleA and angleB` intersect each other at point P. Prove that `angleAPB=90^(@).`

Text Solution

Verified by Experts

AP is the bisector of `angleA.`
`therefore" "angleDAB=2anglePAB" "...(1)`
`"similarly,"" "angleCBA=2anglePBA" "...(2)`
We know that
`angleCBA + angleCBA=180^(@)" "("sum of co-interior angles of a parallelogram is"180^(@))`
`implies" "anglePAB+anglePBA=90^(@)" "["from"(1)and (2)]`
`In angleAPB," "...(3)`
`anglePAB+anglePBA+angleAPB=180^(@)`
`implies" "90^(@)+angleAPB=180^(@)`
`implies" "angleAPB=180^(@)-90^(@)" "("anglesumpoperty")`
`implies" "angleAPB=90^(@)" "["from"(3)]`
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