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

The figure of the question is given below:

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

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)

ABC DCB

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