Q. 33.9( 32 Votes )

In 􀀍ABCD, side BC < side AD (Figure 5.32) side BC || side AD and if side BE side CD then prove that ABC DCB.

Answer :

The figure of the question is given below:

Construction: we will draw a segment to BA meeting BC in E through point D.

Given BC AD

And AB ED (construction)

AB = DE (distance between parallel lines is always same)

Hence ABDE is parallelogram

ABE DEC (corresponding angles on the same side of transversal)

And segBA seg DE (opposite sides of a gram)

But given BA CD

So seg DE seg CD

⇒∠CED DCE ( Δ CED is isosceles with CE = CD)

(Angle opposite to opposite sides are equal)


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In figure 5.23, G is the point of concurrence of medians of ΔDEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that 􀀍GEHF is a parallelogram.

MHB - Math Part-II