In the given figure, ACB = 90° and CD AB. prove that B C ^ 2 A C ^ 2 = B D A D .

In ACD and ABC A = A (Common) ADC = ACB ( each 90°) Thus, By AA similarity criteria ADC ACB Thus, AD AC = AC AB AC^2 = AD ×AB... (i) Similarly, CDB ACB And, DC BC = BC AB BC^2 = DB ×AB... (ii) Dividing (ii) by (i) BC^2 AC^2 = DB AD Hence Proved

In the given figure, ACB = 90° and CD AB. prove that B C ^ 2 A C …

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Question

In the given figure, $\angle$ ACB = 90° and CD $\perp$ AB. prove that $\frac { B C ^ { 2 } } { A C ^ { 2 } } = \frac { B D } { A D }$.

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Answer

In $\triangle $ACD and $\triangle $ ABC $\angle A = \angle A $ (Common) $\angle ADC = \angle ACB$ ( each 90°) Thus, By AA similarity criteria $\triangle ADC \sim \triangle ACB$ Thus, $\frac {AD} {AC} = \frac {AC} {AB} $ $\Rightarrow AC^2 = AD ×AB$... (i) Similarly, $$$\triangle CDB \sim \triangle ACB $ And, $\frac {DC} {BC} = \frac {BC} {AB} $ $\Rightarrow BC^2 = DB ×AB$... (ii) Dividing (ii) by (i) $\frac {BC^2 } {AC^2 } = \frac {DB} {AD} $ Hence Proved $$