In the figure $△ A B C$ is a right angled triangle with right angle at B. BD is perpendicular to AC. Then which of the following options will hold true?

$A B^{2} = A D \times D C$

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$A B^{2} = A D \times A C$

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$A D^{2} = D C \times A C$

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$A B^{2} = D C^{2} + A D^{2}$

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Solution

The correct option is B
$A B^{2} = A D \times A C$

In $△ A B C$ and $△ A D B$
$∠ A B C = ∠ A D B = 90^{o}$
$∠ A = ∠ A$ (common angle)
Therefore, $△ A B C \sim △ A D B$ [by AA similarity]
$\frac{A B}{A D} = \frac{A C}{A B}$
$A B^{2} = A C \times A D$

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Similar questions

In the figure $△ A B C$ is a right angled triangle with right angle at B. BD is perpendicular to AC. Then which of the following options will hold true?

$△$ ABC is a triangle, right angled at B. BD is a perpendicular to AC as shown. Which of the following is true?

Q. In the given figure,

△

A B C

is right angled at

B

. If

B D

⊥

A C

, which of the following is/are true?

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In the figure △ABC is a right angled triangle with right angle at B. BD is perpendicular to AC. Then which of the following options will hold true?

The correct option is B AB2=AD×AC In △ABC and △ADB ∠ABC=∠ADB=90o ∠A=∠A (common angle) Therefore, △ABC∼△ADB [by AA similarity] ABAD=ACAB AB2=AC×AD

In the figure $△ A B C$ is a right angled triangle with right angle at B. BD is perpendicular to AC. Then which of the following options will hold true?

The correct option is B
$A B^{2} = A D \times A C$

In $△ A B C$ and $△ A D B$
$∠ A B C = ∠ A D B = 90^{o}$
$∠ A = ∠ A$ (common angle)
Therefore, $△ A B C \sim △ A D B$ [by AA similarity]
$\frac{A B}{A D} = \frac{A C}{A B}$
$A B^{2} = A C \times A D$