In the figure 6.2

In the figure, we join PA and PD. Draw perpendiculars on AB and CD from P as PM and PN respectively.

AEP = DEP (given)

So, PEN = PEM (M and N are points on line AE and ED respectively)

In ΔPEN and ΔPEM,

PNE = PME = 90°

PEN = PEM

PE = PE (common)

Therefore, Δ PEN ΔPEM (by AAS congruence)

PN = PM (by CPCT)

We know that The chords of a circle equidistant from the center of a circle are congruent.

So, AB = CD.

Hence proved.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Arc of Circle And Related IMP Qs40 mins
IMP Qs Related with Circles For Boards44 mins
Chords Of Circles42 mins
IMP Theorems And Their Application43 mins
Circles- All kinds of Questions36 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 the figure 6.1MHB - Math Part-II

Seg PM and seg PNMHB - Math Part-II

In a circle with MHB - Math Part-II

In the figure 6.2MHB - Math Part-II

Choose correct alMHB - Math Part-II

Choose correct alMHB - Math Part-II

Radius of circle MHB - Math Part-II

In the figure 6.2MHB - Math Part-II