Answer :


Let us drop a perpendicular AG and BH to CD cutting EG at I and J and CD.


In ΔADG & ΔAEI,



AGD = AIE |Right Angle


AEI = ADG |corr. s


ΔADG~ΔAEI by AA Similarity Rule



In ΔBJF & ΔBHC,



BJF = BHC |Right angle


BFJ = BCH |corr. s


ΔBJF~ΔBHC by AA Similarity Rule



In rectangle ABHG & ABJI,


AI = BJ …(a) |opposite sides of rectangle are equal


AG = BH …(b) |opp. sides of rectangle


From eqn. (b) – (a)


AG – AI = BH – BJ


GI = HJ



…(2)


From (1) & (2),



Hence, proved.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
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
caricature
view all courses
RELATED QUESTIONS :

In ΔABC if <span Rajasthan Board Mathematics

In figure PQ and Rajasthan Board Mathematics

You are to selectRajasthan Board Mathematics

The length of theRajasthan Board Mathematics

A give of height Rajasthan Board Mathematics

Figure ΔPRQ ~ ΔTRRajasthan Board Mathematics

If figure, if DE Rajasthan Board Mathematics

The shadow of 12 Rajasthan Board Mathematics

The mid point of Rajasthan Board Mathematics

In figure, if EF Rajasthan Board Mathematics