math-geo-01

# Incircle Contact Chord Meets the Base

In triangle $ABC$ we have $AB = 13$, $BC = 14$, and $CA = 15$.

The incircle of triangle $ABC$ is tangent to side $AB$ at $F$ and tangent to side $CA$ at $E$. Line $EF$ is extended until it meets line $BC$ at a point $P$.

Find the length $BP$.

Give your final answer as an integer.

I need to find the length BP where P is the intersection of line EF with line BC.

First, let me establish coordinates or use properties of the incircle and contact chord.

Let me denote the points where the incircle touches the sides:
- F is where the incircle touches AB
- D is where the incircle touches BC  
- E is where the incircle touches CA

I'll use the property that for a triangle with sides a, b, c, if the incircle touches the sides at points dividing them into segments of lengths x, y, z (where x + y = a, x + z = b, y + z = c), then:
- The semi-perimeter s = (a + b + c)/2
- The tangent segments from each vertex are: s - a, s - b, s - c

For triangle ABC with sides AB = c = 13, BC = a = 14, CA = b = 15:
s = (13 + 14 + 15)/2 = 42/2 = 21

The tangent segments are:
- From A: s - a = 21 - 14 = 7
- From B: s - b = 21 - 15 = 6  
- From C: s - c = 21 - 13 = 8

So:
- AF = AE = 7 (tangent segments from A)
- BF = BD = 6 (tangent segments from B)
- CD = CE = 8 (tangent segments from C)

This means:
- F is on AB with AF = 7, so FB = 13 - 7 = 6
- E is on AC with AE = 7, so EC = 15 - 7 = 8
- D is on BC with BD = 6, so DC = 14 - 6 = 8

Now I need to find where line EF meets line BC at point P.

I'll use the fact that E and F are points on the sides of the triangle, and I can apply Menelaus's theorem or coordinate geometry.

Let me place the triangle in a coordinate system:
- Place B at origin (0, 0)
- Place C on positive x-axis at (14, 0)  
- Find coordinates of A using AB = 13 and AC = 15

If A = (x, y), then:
x² + y² = 169 (distance from B)
(x - 14)² + y² = 225 (distance from C)

From the second equation: x² - 28x + 196 + y² = 225
Substituting the first equation: 169 - 28x + 196 = 225
365 - 28x = 225
28x = 140
x = 5

So y² = 169 - 25 = 144, which gives y = 12.

Therefore: A = (5, 12), B = (0, 0), C = (14, 0)

The points are:
- F is on AB with AF = 7. Since AB = 13, F divides AB in ratio 7:6.
  F = (0 + (7/13)(5-0), 0 + (7/13)(12-0)) = (35/13, 84/13)
  
- E is on AC with AE = 7. Since AC = 15, E divides AC in ratio 7:8.
  E = (5 + (7/15)(14-5), 12 + (7/15)(0-12)) = (5 + 63/15, 12 - 84/15) = (5 + 21/5, 12 - 28/5) = (46/5, 32/5)

Wait, let me reconsider the approach. I'll use a more direct method.

Since F is on AB with AF = 7 and FB = 6, we have F divides AB in ratio 7:6.
Since E is on AC with AE = 7 and EC = 8, we have E divides AC in ratio 7:8.

For the line EF, I can use the fact that P lies on BC (the base) and on line EF.

Using Menelaus's theorem for triangle ABC with transversal EPF:
For triangle ABC and line EF meeting BC at P:
(AF/FB) × (BP/PC) × (CE/EA) = 1

We have:
- AF/FB = 7/6
- CE/EA = 8/7

Substituting into Menelaus's theorem:
(7/6) × (BP/PC) × (8/7) = 1
(56/42) × (BP/PC) = 1
(4/3) × (BP/PC) = 1
BP/PC = 3/4

Since BP + PC = BC = 14:
BP + PC = 14
BP + (4/3)BP = 14
BP(1 + 4/3) = 14
BP(7/3) = 14
BP = 14 × 3/7 = 6

Therefore, BP = 6.

$\boxed{6}$