Sum

In figure, ∠BAC = 90º and segment AD ⊥ BC. Prove that AD^{2} = BD × DC.

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#### Solution

In ∆ABD and ∆ACD, we have

∠ADB = ∠ADC [Each equal to 90º] and, ∠DBA = ∠DAC

[Each equal to complement of angle BAD i.e.,90° - ∠BAD \right]

Therefore, by AA-criterion of similarity, we have

∆DBA ~ ∆DAC

`[ \therefore \ \ \angle D\leftrightarrow \ \angle D,\ \angleDBA\leftrightarrow \ \angle DAC\ and\ \angle BAD\leftrightarrow \ \angleDCA \]`

`\Rightarrow \frac{DB}{DA}=\frac{DA}{DC}`

`\Rightarrow \frac{BD}{AD}=\frac{AD}{DC} `

`AD^2 = BD × DC`

Concept: Similarity

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