The bisectors of $∠ B a n d ∠ C$ of a quadrilateral ABCD intersect in P. Show that P is equidistant from the opposite sides AB and CD.

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Solution

In $Δ P M C$ and $Δ P X C$ ,

$∠ P M C = ∠ P X C$ (perpendiculars)

$∠ P C M = ∠ P C X$ ( bisectors of \angle DCB)

CP = CP(common side)

$∴ Δ P M C ≅ Δ P X C$ (by AAS congruence)

Hence by cpct

PM = PX ........ (1)

Now in $Δ P N B$ and $Δ P X B$ ,

$∠ P N B = ∠ P X B$ (perpendiculars)

$∠ P B N = ∠ P B X$ (bisectors of $∠ C B A$ )

BP = BP(common side)

$∴ Δ P N B ≅ Δ P X B$ (by AAS congruency)

Hence by cpct

PN = PX........ (2)

∴ From (1) and (2) PM = PN

Hence P is equidistant from the opposite sides AB and CD

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The bisectors of ∠Band∠C of a quadrilateral ABCD intersect in P. Show that P is equidistant from the opposite sides AB and CD.

The bisectors of $∠ B a n d ∠ C$ of a quadrilateral ABCD intersect in P. Show that P is equidistant from the opposite sides AB and CD.