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Perpendicular from Right Angle to Hypotenuse Divides the Triangle into Two Similar Triangles

In the figure...

Question

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?

AB² = AD × DC

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AB² = AD × AC

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AD² = DC × AC

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AB² = DC² + AD²

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Solution

The correct option is B
AB² = AD × AC

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
$\frac{A B}{A D} = \frac{A C}{A B}$
$A B^{2} = A C * A D$

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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?

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

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

Q. If ABC is an isosceles triangle and D is a point of BC such that AD ⊥ BC, then

(a) AB² − AD² = BD.DC
(b) AB² − AD² = BD² − DC²
(c) AB² + AD² = BD.DC
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$Δ A B D$ is a right triangle right-angled at A and AC \perp BD. Show that
(i) $A B^{2} = B C . B D$
(ii) $A C^{2} = B C . D C$
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(iv) $\frac{A B^{2}}{A C^{2}} = \frac{B D}{D C}$

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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
AB² = AD × AC

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
$\frac{A B}{A D} = \frac{A C}{A B}$
$A B^{2} = A C * A D$