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
In ΔABC, ∠ACB = 90°
D, E are points on AC, BC. Join AE, DE and BD as shown in the diagram above.
AE2 = AC2 + CE2 (Pythagoras theorem)............... (1)
BD2 = CD2 + BC2 ....................................... (2)
Adding equations (1) and (2)
AE2 + BD2 = AC2 + CE2 + CD2 + BC2
= AC2 + BC2 + EC2 + CD2
= AB2 + DE2 [Since, in ΔABC, AC2 + BC2 = AB2 and in ΔCDE, CD2 + CE2 = DE2]
Therefore, AE2 + BD2 = AB2 + DE2
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
NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.5 Question 13
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. We have proved that AE2 + BD2 = AB2 + DE2.
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