Incentre divides angle bisector ratio proof
Answers
As your question is not clear, I have given the calculation based on my understanding. Hope this helps.
In-centre divides the angle bisectors in some definite ration. Let ABC be a triangle.
AB=b; BC=a; AC=c and let bisector of angle A = AF touching BC at F. Then AF is divided in the ration (b+c) : a. Calculating in this manner for other sides, you will arrive at your answer. The answer should be 11 : 4.
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]
