Q10 of 31 Page 232

P is any a point on diagonal BD of parallelogram ABCD. Let’s prove that triangle APD = triangle CPD.

Given.


P is any a point on diagonal BD of parallelogram ABCD


Formula used.


Area of triangle = × Base × Height



In triangle APD and triangle CPD


As ABCD is a parallelogram


And BD is the diagonal


∴ it divides both congruent triangles


Hence perpendicular of both the triangles are same


AX= CY


For any place of P on BD


The perpendicular of triangle will be same of as of triangle ADB and triangle CBD


Area of triangle APD = × AX × DP


Area of triangle CPD = × CY × DP


= × AX × DP


Area of triangle APD = Area of triangle CPD


∴ triangle APD = triangle APD


More from this chapter

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8

Two triangles ABC and ABD with equal area and stand on the opposite side of AB let’s prove that AB bisects CD.

 9

D is the midpoint of side BC of triangle ABC Parallelogram CDEF stands between side BC and parallel to BC through the point A. Let’s prove that triangle ABC = parallelogram CDEF.

 11

AD and BE are the medians of triangle ABC. Let’s prove that triangle ACD = triangle BCE.

 12

A line parallel to BC of triangle ABC intersects sides AB and AC at the points P and Q respectively. CQ and BQ intersect each other at the point X. Let’s prove that.

(i) ΔBPQ = triangle CPQ


(ii) ΔBCP = triangle BCQ


(iii) ΔACP = triangle ABQ


(iv) ΔBXP = triangle CXQ