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#### Question

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

#### Solution

#### Similar questions

Prove that the diagonals of a rectangle ABCD, with vertices A(2, -1), B(5, -1), C(5, 6) and D(2, 6), are equal and bisect each other.

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In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

(i) OA^{2} + OB^{2} + OC^{2} − OD^{2} − OE^{2} − OF^{2} = AF^{2} + BD^{2} + CE^{2}

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