In the given figure,ABC is a triangle with ∠B = 90° , BC = 5 cm and AB = 12 cm.
D is point on AC such that AD = 4 cm and E is the midpoint of AB. Join D and E and extend DE to meet CB at F. Find BF.

3 cm

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6 cm

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5 cm

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4 cm

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Solution

The correct option is D 4 cm

Consider $Δ A B C,$
D, E and F are respective points on the sides CA, AB and BC. By construction D, E, F are collinear.
By Menelaus’ theorem
$\frac{A E}{E B} \times \frac{B F}{F C} \times \frac{C D}{D A} = 1$ ... (1)
By assumption, AE = EB = 6 cm, DA = 4 cm and
FC = FB + BC = BF + 5
By Pythagoras theorem, $(A C)^{2} = (A B)^{2} + (B C)^{2} = 144 + 25 = 169,$
Therefore AC = 13 cm
and So, CD = AC – AD = 13 – 4 = 9 cm.
Substituting the values of FC, AE, EB, DA, CD in (1),
we get,
$\frac{6}{6} \times \frac{B F}{B F + 5} \times \frac{9}{4} = 1$
$\Rightarrow$ 9BF = 4BF + 20
$∴ B F = 4 c m .$

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