In the given figure, AD ⊥ BC and BD = 1/3 CD. Prove that: 2CA² = 2AB² + BC².

Question

In the given figure, AD ⊥ BC and BD = 1/3 CD. Prove that: 2CA 2 = 2AB 2 + BC 2 .

Philoid · Accepted Answer

We have,

BC = BD + CD

[1]

As, AD ⏊ BC

⇒ ΔADC is a right-angled triangle

By Pythagoras theorem,[i.e. hypotenuse² = perpendicular² + base²]

AD² + CD²= CA²

⇒ AD² = CA² - CD² ….[2]

Also, ΔABD is a right-angled triangle

By Pythagoras theorem,

AD² + BD² = AB²

From [2]

CA² - CD² + BD² = AB²

⇒

[From [1]]

⇒ 2CA² - BC² = 2AB²

⇒ 2CA² = 2AB² + BC²

Hence, Proved.