Question

Let ABC be an isosceles triangle with AB=AC and incentre I. If AI = 3 and the distance from I to BC is 2, what is the square of the length of BC? ​

Answer 1

Answer:

GIVEN: A triangle ABC, AI, BI & CI are angle bisectors. I is incentre of the triangle. BC =a, AC = b, & AB = c

TO PROVE: AI/ID = (b+c)/a

PROOF: By applying angle bisector theorem: which states that: Angle bisector of any angle of a triangle, bisects the opposite side in the same ratio, as that of the other 2 sides of the triangle.

So, in triangle ABD,

BI bisects angle B ( given)

=> c/ BD = AI/ID ( by angle bisector theorem) …………………..(1)

Now, in triangle ACD,

CI bisects angle C ( given)

=> b/DC = AI/ ID ( by the above theorem) …………………..(2)

By (1) & (2)

c/ BD = b/ DC

=> DC/BD = b/c

=> (DC + BD)/BD = ( b+c) /c ( by adding 1 to both the sides or by law of proportion)

=> a/ BD = ( b+ c)/c

=> c/BD = (b+c)/a

But c/BD = AI/ID ( by (1) )

Hence, AI/ ID = (b+c)/a

[ Hence Proved]

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