Question

d and E are the points on the sides CA and CB respectively of a triangle ABC right angled at C prove that AE²+BD²=AB²+DE²​

Answer 1

Given:              A right triangle ABC, right angled at C. D and E are

                       points on side AC and BC respectively.

To prove:         AE^2+BD^2=AB^2+DE^2

Construction: Join AE, BC and DE

Proof:               In triangle ACE

                       AE^2=AC^2+CE^2___________(I)

                       [By Pythagorean’s theorem]

                       In triangle BCD

                      BD^2=CD^2+BC^2___________(Il)

                      [By Pythagorean’s theorem]

                     Add (l) and (ll) we get,

                     AE^2+BD^2=(AC^2+BC^2) + (CE^2+CD^2)

                     AE^2+BD^2=AB^2+DE^2

                     Hence proved                              

Answer 2

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