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.

The side lengths of triangle $ABC$ are $c = AB = 13$, $a = BC = 14$, and $b = CA = 15$.
First, we calculate the semi-perimeter $s$:
$$s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21$$

The distances from the vertices to the points of tangency of the incircle are given by $s$ minus the length of the opposite side.
Let $D, E, F$ be the points of tangency on $BC, CA, AB$ respectively.
The distance from vertex $A$ to the points of tangency on adjacent sides $AB$ and $AC$ is:
$$AF = AE = s - a = 21 - 14 = 7$$
The distance from vertex $B$ to the point of tangency on $AB$ is:
$$BF = s - b = 21 - 15 = 6$$
The distance from vertex $C$ to the point of tangency on $AC$ is:
$$CE = s - c = 21 - 13 = 8$$
(Check: $AF + FB = 7 + 6 = 13 = AB$, $AE + EC = 7 + 8 = 15 = AC$. Correct.)

We are considering the line passing through $E$ and $F$, which intersects the line $BC$ at point $P$. We can apply Menelaus' Theorem to triangle $ABC$ and the transversal line $P-E-F$. The theorem states:
$$\frac{AF}{FB} \cdot \frac{BP}{PC} \cdot \frac{CE}{EA} = 1$$
Substituting the known lengths:
$$\frac{7}{6} \cdot \frac{BP}{PC} \cdot \frac{8}{7} = 1$$
Simplifying the equation:
$$\frac{7}{6} \cdot \frac{8}{7} \cdot \frac{BP}{PC} = 1$$
$$\frac{8}{6} \cdot \frac{BP}{PC} = 1$$
$$\frac{4}{3} \cdot \frac{BP}{PC} = 1$$
$$\frac{BP}{PC} = \frac{3}{4}$$

Since the line $EF$ intersects the sides $AB$ and $AC$ internally (at points $F$ and $E$), it must intersect the line containing the third side $BC$ externally. Thus, $P$ lies outside the segment $BC$.
The ratio $BP/PC = 3/4$ implies that $BP < PC$. Therefore, $P$ is closer to $B$ than to $C$, which means $P$ lies on the extension of $CB$ past $B$.
Let $BP = x$. Then $PC = PB + BC = x + 14$.
Substituting into the ratio:
$$\frac{x}{x + 14} = \frac{3}{4}$$
$$4x = 3(x + 14)$$
$$4x = 3x + 42$$
$$x = 42$$

Thus, the length $BP$ is 42.

\boxed{42}