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# D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.Q.13

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Hello, sir please help me the best way to solve this problem of class 10th chapter of triangles chapter of exercise 6.5 of question no.13 give me the simple way to solve this problem, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.

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1. Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

By Pythagoras theorem in Î”ACE, we get

AC2Â +Â CE2Â = AE2Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(i)

In Î”BCD, by Pythagoras theorem, we get

BC2Â +Â CD2Â = BD2Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

From equationsÂ (i)Â andÂ (ii), we get,

AC2Â +Â CE2Â + BC2Â +Â CD2Â = AE2Â + BD2Â â€¦â€¦â€¦â€¦..(iii)

In Î”CDE, by Pythagoras theorem, we get

DE2Â =Â CD2Â + CE2

In Î”ABC, by Pythagoras theorem, we get

AB2Â =Â AC2Â + CB2

Putting the above two values in equationÂ (iii), we get

DE2Â + AB2Â = AE2Â + BD2.

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