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)


ABC DCB


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Quiz | Properties of Parallelogram31 mins
Critical Thinking Problems on Quadrilaterals44 mins
Quiz | Basics of Quadrilaterals36 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

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