If the bisector of angles A and B of a quadrilateral ABCD intersect each other at the point P. The point P is equidistant from AD and BC.

True

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False

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Solution

The correct option is A True

From P, perpendiculars PL on AD, PM on AB and PN on BC respectively are drawn.
P lies on the bisector of angle A, hence, PL = PM -------------------- (i)
Again, P lies on bisector of angle B, PM = PN ------------------------ (ii)
From (i) and (ii) we get, PL = PN i.e. P is equidistant from AD and BC. [Proved.]

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