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Basic Proportionality Theorem

In Δ ABC Fig ...

Question

In $Δ A B C$ (Fig 7.60), if $∠ 1 = ∠ 2$ , prove that $\frac{A B}{A C} = \frac{B D}{D C}$ .

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Solution

for ADB and ADC
BAD = CAD (given)
ADB =ADC ( right angle)
AD = AD (common)
Both the triangles are similar by ASA
Hence,
$fraction numerator A B over denominator A C end fraction equals fraction numerator B D over denominator C D end fraction$

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Similar questions

In Fig 7.60, AD bisects $∠$ A, AB = 12 cm, AC = 420 cm and BD = 5 cm determine CD.

△

ABC ( Fig.4.59 ), if

∠

∠

\frac{A B}{A C} = \frac{B D}{D C}

.

A C = B D,

then prove that

A B = C D

.

△ A B C, A B = A C, B D ⊥ A C

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

Δ A B C, A B = A C

B D ⊥ A C

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

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