Q17 of 31 Page 232

ABCD is a parallelogram and ABCE is a quadrilateral shaped regions. Diagonal AC divides the quadrilateral shaped region ABCE into two equal parts. Let’s prove that AC || DE.

Given.


ABCD is a parallelogram and ABCE is a quadrilateral


Diagonal AC divides the quadrilateral shaped region ABCE into two equal parts


Formula used.


If 2 triangles are on same base And having equal area then they lies between 2 parallel lines.


Diagonal of parallelogram divide it into 2 congruent triangles



In quadrilateral ABCE


As AC divides quadrilateral into 2 equal parts


Triangle ABC = triangle AEC


In parallelogram ABCD


As AC is diagonal which divides parallelogram into 2 equal triangles


Triangle ABC = triangle ADC


Comparing both


We get


Triangle AEC = triangle ADC


As both triangles triangle AEC ,Δ ADC are equal


And both lies on same base AC


Hence both triangles comes between parallel lines AC , DE


∴ AC || DE


More from this chapter

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15

In triangle ABC ∠ABC = ∠ACB; bisectors of an angle ∠ABC and ∠ACB intersect the side AC and AB at the points E and F respectively. Let’s prove that, FE || BC.

 16

The two parallelogram shaped regions ABCD and AEFG of which ∠A is common are equal in area and E lies on AB. Let’s prove that DE || FC.

 18

D is the midpoint of side BC of triangle ABC; P and Q lie on sides BC and BA in such a way that triangle BPQ = ΔABC. Let’s prove that, DQ || PA.

 19

Parallelogram ABCD of which midpoints are E, F, G and H of sides AB, BC, CD and DA respectively. Let’s prove that

(i) EFGH is a parallelogram.


(ii) Area of parallelogram shaped region EFGH is half of the area of parallelogram shaped region ABCD.