#### Question

In figure, ∠B of ∆ABC is an acute angle and AD ⊥ BC, prove that AC^{2} = AB^{2} + BC^{2} – 2BC × BD

#### Solution

Given: A ∆ABC in which ∠B is an acute angle and AD ⊥ BC.

To Prove: AC^{2} = AB^{2} + BC^{2} – 2BC × BD.

Proof: Since ∆ADB is a right triangle right-angled at D. So, by Pythagoras theorem, we have

AB^{2} = AD^{2} + BD^{2} ….(i)

Again ∆ADC is a right triangle right angled at D.

So, by Pythagoras theorem, we have

AC^{2} = AD^{2} + DC^{2}

⇒ AC^{2} = AD^{2} + (BC – BD)^{2}

⇒ AC^{2} = AD^{2} + (BC^{2} + BD^{2} – 2BC • BD)

⇒ AC^{2} = (AD^{2} + BD^{2} ) + BC^{2} – 2BC • BD

⇒ AC^{2} = AB^{2} + BC^{2} – 2BC • BD [Using (i)]

Hence, AC^{2} = AB^{2} + BC^{2} – 2BC • BD

Is there an error in this question or solution?

Solution In figure, ∠B of ∆ABC is an acute angle and AD ⊥ BC, prove that AC^2 = AB^2 + BC^2 – 2BC × BD Concept: Pythagoras Theorem.