5.
In figure, P is a point in the interior of a parallelogram ABCD. Show that:
(1) ar(
(ii) ar(ΔAPD) + ar(ΔPBC) = ar(ΔAPB) + ar(ΔPCD). [CBSE 2012 (March)]
[Hint. Through P, draw a line parallel to AB.]
Given: P is a point in the interior of a parallelogram ABCD.
To Prove : (i)
ar(|| gm ABCD)
(ii) ar(ΔAPD) + ar(ΔPBC) = ar(ΔAPB) + ar(ΔPCD).
Construction: Through P, draw a line EF parallel to AB.
Proof: (i) EF || AB ...(1) | by construction
∵ AD || BC
∵ Opposite sides of a parallelogram are parallel
∴ AE || BF ...(2)
In view of (1) and (2),
Quadrilateral ABFE is a parallelogram
A quadrilateral is a parallelogram if its opposite sides are parallel
Similarly, quadrilateral CDEF is a parallelogram
∵ ΔAPB and || gm ABFE are on the same base AB and between the same parallels AB and EF.
∵ ΔPCD and || gm CDEF are on the same base DC and between the same parallels DC and EF.Adding (3) and (4), we get ar(ΔAPB) + ar(ΔPCD)
(ii) ar(ΔAPD) + ar(ΔPBC)
= ar(|| gm ABCD) – [ar(ΔAPB) + ar(ΔPCD)] = 2 [ar(ΔAPB) + ar(ΔPCB)] – [ar(ΔAPB) + ar(ΔPCD)]
= ar(ΔAPB) + ar(ΔPCD).
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