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Perpendicular from right angle to hypotenuse divides the triangle into two similar triangles

In a right AB...

Question

In a right $△$ ABC, a perpendicular BD is drawn on to the largest side from the opposite vertex. Which of the following does not give the ratio of the areas of $△$ ABD and $△$ ACB?

$({\frac{A B}{A C})}^{2}$

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$({\frac{A D}{A B})}^{2}$

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$({\frac{B D}{C B})}^{2}$

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$({\frac{A B}{A D})}^{2}$

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Solution

The correct option is D
$({\frac{A B}{A D})}^{2}$

Consider ΔABD and ΔACB:

∠BAD = ∠BAC [common angle]

∠BDA = ∠ABC [ $90^{\circ}$ ]

By AA similarity criterion,
$△ A B D ~ △$ ACB

Hence,

$\frac{a r (Δ A B D)}{a r (Δ A C B)} = ({\frac{A B}{A C})}^{2} = ({\frac{A D}{A B})}^{2} = ({\frac{B D}{C B})}^{2}$

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In a right △ABC, a perpendicular BD is drawn on to the largest side from the opposite vertex. Which of the following does not give the ratio of the areas of △ABD and △ACB?

The correct option is D (ABAD)2 Consider ΔABD and ΔACB: ∠BAD = ∠BAC [common angle] ∠BDA = ∠ABC [ 90∘] By AA similarity criterion, △ ABD ~ △ACB Hence, ar(ΔABD)ar(ΔACB)=(ABAC)2=(ADAB)2=(BDCB)2

In a right $△$ ABC, a perpendicular BD is drawn on to the largest side from the opposite vertex. Which of the following does not give the ratio of the areas of $△$ ABD and $△$ ACB?