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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# 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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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

AC

^{2}Â +^{Â }CE^{2}Â = AE^{2}Â 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

BC

^{2}Â +^{Â }CD^{2}Â = BD^{2}Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(ii)From equationsÂ

(i)Â andÂ(ii), we get,AC

^{2}Â +^{Â }CE^{2}Â + BC^{2}Â +^{Â }CD^{2}Â = AE^{2}Â + BD^{2}Â â€¦â€¦â€¦â€¦..(iii)In Î”CDE, by Pythagoras theorem, we get

DE

^{2}Â =^{Â }CD^{2}Â + CE^{2}In Î”ABC, by Pythagoras theorem, we get

AB

^{2}Â =^{Â }AC^{2}Â + CB^{2}Putting the above two values in equationÂ

(iii), we getDE

^{2}Â + AB^{2}Â = AE^{2}Â + BD^{2}.