Question

D and e are the points on sides CA and CB respectively of triangle ABC right angled at C prove that AE^2 +BD^2=AB^2+DE^2
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Answer 1

Step-by-step explanation:

In triangle ACE,

AC^2 + CE^2 = AE^2 ...(i)    

(By PythagorasTheorem)

In triangle DBC,

DC^2 + BC^2 = BD^2 ...(ii)  

  (By PythagorasTheorem)

In triangle ABC,

AC^2 + BC^2 = AB^2 ...(iii)  

  (By Pythagoras Theorem)

In triangle DEC,

DC^2 + CE^2 = DE^2 ...(iv)     

(By Pythagoras Theorem)

Adding (i) and (ii) we get,

AE^2 + BD^2 = AC^2 + CE^2+ DC^2+ BC^2

AE^2 +BD^2 = AC^2 +BC^2 +CE^2 + DC^2

AE^2 + BD^2 = AB^2 + DE°2                        [ from (iii) and (iv)

hence prove that

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Answer 2

hope it's help u........