Question

✴️AhôLá U$èRs✴️
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√√Here is your question√√

D and E are points on the side CA and CB respectively of a triangle ABC right angled at C .Prove that
AE² + BD² = AB² + DE².

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Answer 1

In triangle ACE,


AC2 ​+ CE2 = AE2 ...(i)    (By Pythagoras Theorem)
In triangle DBC,


DC2 ​+ BC2 = BD2 ...(ii)    (By Pythagoras Theorem)


In triangle ABC,
AC2 + BC2 = AB2 ...(iii)    (By Pythagoras Theorem)


In triangle DEC,
DC2 + CE2 = DE2 ...(iv)     (By Pythagoras Theorem)​

Adding (i) and (ii) we get,
AE2 + BD2 = AC2 + CE2+ DC2+ BC2
AE2 +BD2 = AC2 +BC2 +CE2 + DC2
AE2 + BD2 = AB2 + DE2                                 [ from (iii) and (iv)]


HENCE PROVED ♥️


              

$hope \: helps$

Answer 2

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