Question

In the given figure, AD ⊥ BC and BD = 1/3 CD. Prove that : CA2 = AB2 + BC2.

Answer 1

Sure

Could you mark me as brainliest

∵ AD ⊥ BC. ∴∠ADB = ∠ADC = 90°. In right angled triangle ADC, CA2 = CD2 + AD2 … (1) (By using Pythagoras theorem) Also, In triangle ADB, AB2= BD2+AD2 (Using Pythagoras theorem) ⇒ AD2 = AB2 − BD2 Now, Putting value of AD2 in equation (1), we get CA2 = CD2 + AB2 − BD2 = 9BD2 + AB2 − BD2 (∵ BD = 1 3 13CD (given) ⇒ CD = 3BD) = 8BD2 + AB2 = 8 ( 1 4 14BC)2 + AB2 (∵ BC = BD + CD = BD+ 3BD = 4BD ⇒ BD = 1 4 14BC) = 8 16 816BC2+ AB2 = 1 2 12BC2 + AB2 ⇒ 2CA2 = BC2 + 2AB2 Hence Proved.