# In triangle ABC, it is known that AB = 15, BC = 14, AC = 13, and the median AA 1

In triangle ABC, it is known that AB = 15, BC = 14, AC = 13, and the median AA 1 intersects the bisector BB1 at point P, find the area A1PB1C.

Let us define the area of ​​the triangle ABC by Heron’s theorem.

The semi-perimeter of the triangle ABC is equal to: p = (AB + BC + AC) / 2 = 42/2 = 21 cm.

Then Saс = √21 * (21 – 15) * (21 – 14) * (21 – 13) = √7056 = 84 cm2.

Since AA1 is the median of the ABC triangle, then Saa1b = Saa1s = Saabs / 2 = 42 cm2.

The segment BB1 is the bisector of triangles ABC and AA1B, then AB / BC = AB1 / CB1 = 15/14.

Then Svsv1 / Sawv1 = 15/14.

Svsv1 = 15 * Savv1 / 14.

Svsv1 + Savv1 = Savs = 84.

15 * Savv1 / 14 + Savv1 = 84.

29 * Savv1 / 14 = 84.

Savv1 = 1176/29 = 40.55 cm2.

Similarly:

AB / BA1 = AP / A1P = 15/7.

Then Saur / Sa1vr = 15/7.

Sa1vr = 7 * Saur / 15.

Savr + Savr = Sava1 = 42.

Savr + 7 * Savr / 15 = 42.

22 * Savr / 15 = 42.

Sarv = 630/22 = 28.64 cm2.

Then Sarv1 = Saav1 – Sarv = 40.55 – 28.64 = 11.91 cm2.

Then Sa1rv1s = Saa1s – Saav1 = 42 – 11.91 = 30.09 cm2.