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Question

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.

A
3 cm
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B
6 cm
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C
5 cm
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D
4 cm
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Solution

The correct option is D 4 cm

Consider ΔABC,
D, E and F are respective points on the sides CA, AB and BC. By construction D, E, F are collinear.
By Menelaus’ theorem
AEEB×BFFC×CDDA=1 ... (1)
By assumption, AE = EB = 6 cm, DA = 4 cm and
FC = FB + BC = BF + 5
By Pythagoras theorem, (AC)2=(AB)2+(BC)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,
66×BFBF+5×94=1
9BF = 4BF + 20
BF=4 cm.

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