Q8 of 31 Page 232

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

Given.


Two triangles ABC and ABD with equal area and stand on the opposite side of AB


Formula used.


Area of triangle = × Base × Height


AAS congruency rule = If 2 angles and one side of both triangles are equal then both triangle are congruent.


Make the perpendicular CE and DF on triangle ABC and triangle ABD


Mark intersection point of AB and CD as O



If Area of triangle ABC = Area of triangle ABD


× AB × CE = × AB × DF


CE = DF


In triangle CEO and triangle DFO


∠ DFO = ∠ CEO ∵ Both are perpendicular 90°


∠ DOF = ∠ COE ∵ vertically opposite angle


CE = DF ∵ proved above


triangle CEO ≅ triangle DFO ∵ AAS congruency rule


∴ CO = DO


Hence AB bisect CD


More from this chapter

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6

O is any point outside the equilateral triangle ABC and within the angular region on ABC; OP, OQ and OR are the perpendicular distance of AB, BC and CA respectively from the point O. Let us prove that the altitude of the triangle = OP + OQ – OR.

 7

E is any point on side DC of parallelogram ABCD, produced AE meets produced BC at the point F. Joined D, F. Let’s prove that

(i) triangle ADF = triangle ABE


(ii) triangle DEF = triangle BEC.

 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.

 10

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