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Pythagoras Theorem

In a right an...

Question

In a right angled $△$ ABC , AC $⊥$ the hypotenuse BC then AD $^{2}$ is

A C \times C D

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A C \times B A

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D C \times B D

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None of these

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Solution

The correct option is B $D C \times B D$
From the figure, it is clear that $Δ C A D s i m i l a r t o Δ D A B b y A A A$ .
So, $\frac{B D}{A D} = \frac{A D}{C D} \to {A D}^{2} = B D * C D$

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Similar questions

In $△ A B C$ , B is the right angle and BD is perpendicular to AC, then:

A B C

B D

A C

(i) A C \times A D = A B^{2}

(i i) A C \times C D = B C^{2}

A B D

A

A C

⊥

A B^{2} = B C . B D

A C^{2} = B C . D C

A D^{2} = B D . C D

\frac{{AB}^{2}}{{AC}^{2}} = \frac{BD}{DC}

△ A B C, ∠ A

A D ⊥ B C

D ϵ B C

A D^{2} = B D \cdot D C

B C^{2} = A D \cdot D C

A B^{2} = B D \cdot B C

A C^{2} = C D \cdot B D

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