D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE^2 + BD^2 = AB^2 + DE^2.
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06/06/2021 12:10 pm
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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.
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06/06/2021 12:11 pm
By Pythagoras theorem in ΔACE, we get
AC2 + CE2 = AE2 …………….(i)
In ΔBCD, by Pythagoras theorem, we get
BC2 + CD2 = BD2 ………………..(ii)
From equations (i) and (ii), we get,
AC2 + CE2 + BC2 + CD2 = AE2 + BD2 …………..(iii)
In ΔCDE, by Pythagoras theorem, we get
DE2 = CD2 + CE2
In ΔABC, by Pythagoras theorem, we get
AB2 = AC2 + CB2
Putting the above two values in equation (iii), we get
DE2 + AB2 = AE2 + BD2.
This post was modified 3 years ago by Raavi Tiwari
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