Question

D and e are the points on sides ca and cb respectively of triangle abc right angle at c provr that ae square +bd square =ab square + de square

Answer 1

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Answer 2

Hence, proved that AE^2 + BD^2 = AB^2 + DE^2.

Step-by-step explanation:

Given that,

△ABC is right angled at C

D and E are the two points drawn on sides AC and AB

To prove,

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

By joining the point A with E and B with D.

Using Pythagoras theorem,

H^2 = P^2 + B^2

In △ACE,

AE^2 = AC^2 + CE^2 ...(i)

In △BCD,

BD ^2  = DC^2  + BC^2  ...(ii)

In △ABC,

AB^2 = AC^2 + BC^2  ...(iii)

In △DCE,

DE^2 = DC^2 + CE^2 ...(iv)

Thus,

AC^2 + CE^2 + DC^2  + BC^2 = AC^2 + BC^2 + DC^2 + CE^2

L.H.S = R.H.S

∵  AE^2 + BD^2 = AB^2 + DE^2

Learn more: Prove that AE^2 + BD^2 = AB^2 + DE^2

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