KVPY SA Exam 2014 Question Paper

Question : 8 of 80

Marks: +1, -0

The angle bisectors BD and CE of a triangle ABC are divided by the incentre I in the ratios 3 :2 and 2:1 respectively. Then the ratio in which I divides the angle bisector through A is

3 : 1
11 : 4
6 : 5
7 : 4

Validate

Solution:

I divides AF
b + c : a
I divides BD
c + a : b
I divides CE in
a + b : c

{c+a}/b=3/2;{a+b}/c=2/1

2c + 2a = 3b

⇒ a + b + 2a = 3b

⇒ 3a = 2b

⇒

b={3a}/2

⇒ a + b = 2c

a +{3a}/2=2c

5/2a=2c

c=5/4a

Now

{b+c}/a={{3a}/2+{5a}/4}/a

={6+5}/4=11/4

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