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 AE^{2} + BD^{2} = AB^{2} + DE^{2}.
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06/06/2021 12:11 pm
By Pythagoras theorem in ΔACE, we get
AC^{2} +^{ }CE^{2} = AE^{2} …………….(i)
In ΔBCD, by Pythagoras theorem, we get
BC^{2} +^{ }CD^{2} = BD^{2} ………………..(ii)
From equations (i) and (ii), we get,
AC^{2} +^{ }CE^{2} + BC^{2} +^{ }CD^{2} = AE^{2} + BD^{2} …………..(iii)
In ΔCDE, by Pythagoras theorem, we get
DE^{2} =^{ }CD^{2} + CE^{2}
In ΔABC, by Pythagoras theorem, we get
AB^{2} =^{ }AC^{2} + CB^{2}
Putting the above two values in equation (iii), we get
DE^{2} + AB^{2} = AE^{2} + BD^{2}.
This post was modified 3 years ago by Raavi Tiwari
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