In the given figure, ABC is a triangle with ∠B = 90° , BC = 3 cm and AB = 4 cm.
D is point on AC such that AD = 1 cm and E is the midpoint of AB. Join D and E and extend DE to meet CB at F. Find BF.

4 cm

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3 cm

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1 cm

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2 cm

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Solution

The correct option is C 1 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 = 2 cm, DA = 1 cm and
FC = FB + BC = BF + 3
By Pythagoras theorem, $(A C)^{2} = (A B)^{2} + (B C)^{2} = 16 + 9 = 25,$
Therefore AC = 5 cm
and So, CD = AC – AD = 5 – 1 = 4 cm.
Substituting the values of FC, AE, EB, DA, CD in (1),
we get,
$\frac{2}{2} \times \frac{B F}{B F + 3} \times \frac{4}{1} = 1$
$\Rightarrow$ 4BF = BF + 3
$∴ B F = 1 c m .$

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