Question

In ∆ ABC, ∠B = 90°,BD ⊥ AC then Prove that AB2 = AC × AD​

Answer 1

Answer:

${ab}^{2} = bc \times bd \: \: {[using \: the \: corollary]} \\ { \sqrt[2]{5} }^{2} = bc \times 4 \\ = \cancel\dfrac{20}{4} = bc \\ 5 = bc \\ now \: cd = bc - bd \\ = 5 - 4 \\ = 1 \\ {ac}^{2} = bc \times bd \\ ac = 5 \times 1 \\ ac = \sqrt{5}$

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