Sum

In Fig., ∆ABC is an obtuse triangle, obtuse angled at B. If AD ⊥ CB, prove that AC^{2} = AB^{2} + BC^{2} + 2BC × BD

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#### Solution

**Given**: An obtuse triangle ABC, obtuse-angled at B and AD is perpendicular to CB produced.

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

Proof: Since ∆ADB is a right triangle right angled at D. Therefore, by

Pythagoras theorem, we have AB^{2} = AD^{2} + DB^{2} ….(i)

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

Therefore, by Phythagoras theorem, we have

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

⇒ AC^{2} = AD^{2} + (DB + BC)^{2}

⇒ AC^{2} = AD^{2} + DB^{2} + BC^{2} + 2BC • BD

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

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

Concept: Right-angled Triangles and Pythagoras Property

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