" triangle ABC and I is a point in triangle ABC such that angle IBA > angle ICA , angle IBC > angle ICB , BI cut AC at B' , CI cut AB at C' . PRove the : BB' < CC' "

So there is a triangle ABC.

1) A line segment from corner B is drawn touching side AC at B'.

2) A line segment from corner C is drawn touching side AB at C'.

3) The two line segments, BB' and CC', intersect at point I.

4) Angle IBA (and so, angle B'BA) is greater than angle ICA (and so, angle C'CA).

5) Angle IBC (and so, angle B'BC) is greater than angle ICB (and so, angle C'CB).

Prove line segment BB' is shorter than line segment CC'.

Okay.

Combining or from 4) and 5), angle ABC is greater than angle BCA.

Hence, side AC is greater than side AB.

Then, area of triangle ABC = area of triangle ABC. [ :-) ]

Area of any triangle is (base)*(altitude)/2.

If base is side AC, area of triangle ABC = (AC)*(altitude from AC to apex B)/2.

This altitude, say, h1, is proportional to line segment BB'.

If base is side AB, area of triangle ABC = (AB)*(altitude from AB to apex C)/2.

This altitude, say, h2, is proportional to line segment CC'.

h1 is shorter than h2.

Because, with area of triangle ABC being constant or not changing, and AC being longer than AB, then h1 must be shorter than h2.

Hence, by proportion, BB' is shorter than CC'. --------------****

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In triangle IBC, IB is shorter than IC, because angle ICB is less than angle IBC.

But I cannot connect that fact with the other fact that AB is shorter than AC.