In a right-angled triangle with sides $a$ and $b$ and hypotenuse $c$, the altitude drawn on the hypotenuse is $x$. Prove that $ab\ =\ cx$.

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**Given:**

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In a right-angled triangle with sides $a$ and $b$ and hypotenuse $c$, the altitude drawn on the hypotenuse is $x$.

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**To do:**

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We have to prove that $ab\ =\ cx$. ”Š

**Solution:**

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In the given figure, $\vartriangle ABC$ is a right angle triangle having sides $a$ and $b$ and hypotenuse $c$.

Let $BD$ be the altitude drawn on the hypotenuse $AC$.

In $\vartriangle ACB$ and $\vartriangle CDB$,

$\angle B = \angle B$ (Common)

$\angle ACB = \angle CDB = 90^o$

Therefore,

$\vartriangle ACB ∠¼ \vartriangle CDB$ (By AA similarity)

Hence,

$\frac{AB}{BD} = \frac{AC}{BC}$ (Corresponding parts of similar triangles are proportional)

$\frac{a}{x} = \frac{c}{b}$

$ab = cx$

Hence proved.