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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Solution

Applying Pythagoras theorem in ΔACE, we obtain

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Q. Question 13
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that

A E^{2} + B D^{2} = A B^{2} + D E^{2}

D

and

E

are points on the sides

C A

and

C B

repesctively of a

△ A B C

, right angled at

C

. Prove that

A E^{2} + B D^{2} = A B^{2} + D E^{2}

Q. In given figure ABC is a right-angled triangle at B. Let D and E be any points on AB and BC respectively. Prove that

{A E}^{2} + {C D}^{2} = {A C}^{2} + {D E}^{2}

Q. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Then:

P and Q are points on sides CA and CB respectively of Δ ABC right angled at C. Prove that ${A Q}^{2}$ + ${B P}^{2}$ = ${A B}^{2}$ + ${P Q}^{2}$ [4 MARKS]

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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

Applying Pythagoras theorem in ΔACE, we obtain

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