Question

In triangleABC, BE and CF are the angular bisectors of angleB and angleC meeting at I. Prove that
AF/FI = AC/CI.
pls solve but if you dont know please dont solve​

Answer 1

Step-by-step explanation:

Let ABC be a triangle. Let BE and CF be internal angle bisectors of ∠B and ∠C, respectively, with E on AC and F on AB. Suppose X is a point on the segment CF such that AX⊥CF, and Y is a point on the segment BE such that AY⊥BE. 

Prove that XY=2(b+c−a), where BC=a, CA=b, and AB=c.

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Produce AX and AY to meet BC is X' and Y' respectively.

Since BY bisects ∠ABY′  and BY⊥AY′ it follows that BA = BY' and AY = YY'.

Similarly, CA=CX′ and AX=XX′.

Thus X and Y are mid-points of AX' and AY' respectively. By mid-point theorem XY = X'Y'/2.  But X′Y′=X′C+Y′B−BC=AC+AB−BC=b+c−a

Hence XY=(b+c−a)/2.