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Question
The bisectors of ∠B and ∠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
Given: A quadrilateral ABCD in which bisectors of ∠B and ∠C meet in P. PM ⊥ ABand PN ⊥ CD.
To prove: PM = PN
Construction: Draw PL ⊥ BC
Proof: Since, P lies on the bisector of ∠B
∴ P is equidistant from BC and BA
⇒ PL = PM ...(i)
Also, P lies on the bisector of ∠C ...[Given]
∴ P is equidistant from CB and CD
⇒ PL = PN ...(ii)
From (i) and (ii), we have
PL = PM
and PL = PN
⇒ PM = PN.
Hence proved.
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