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.

Let the vertices of the triangle be $A, B, C$.
The side lengths are given as $c = AB = 13$, $a = BC = 14$, $b = CA = 15$.
Let $I$ be the incenter of $\triangle ABC$.
The incircle is tangent to $AB$ at $F$ and to $CA$ at $E$.
We need to find the length $BP$, where $P$ is the intersection of line $EF$ and line $BC$.

First, let's determine the lengths of the segments from the vertices to the points of tangency.
Let $s$ be the semi-perimeter of $\triangle ABC$.
$s = \frac{a+b+c}{2} = \frac{14+15+13}{2} = \frac{42}{2} = 21$.
The distance from a vertex to the points of tangency on the adjacent sides is given by $s - \text{opposite side}$.
Let $D$ be the point of tangency on $BC$.
Then $AF = AE = s - a = 21 - 14 = 7$.
$BF = BD = s - b = 21 - 15 = 6$.
$CD = CE = s - c = 21 - 13 = 8$.
Let's check if these sum up correctly.
$AB = AF + FB = 7 + 6 = 13$. Correct.
$BC = BD + DC = 6 + 8 = 14$. Correct.
$CA = CE + EA = 8 + 7 = 15$. Correct.

So, $E$ is on $AC$ such that $AE = 7$ and $EC = 8$.
$F$ is on $AB$ such that $AF = 7$ and $FB = 6$.
The line $EF$ intersects $BC$ at $P$.
We need to find $BP$. Let $BP = x$.
Since $P$ is on the line $BC$, $P$ could be on the segment $BC$ or outside.
Let's visualize the configuration.
$A$ is the top vertex. $B$ is bottom left, $C$ is bottom right.
$F$ is on $AB$, closer to $B$ than $A$? No, $AF=7, FB=6$. $AB=13$. So $F$ is roughly in the middle.
$E$ is on $AC$, closer to $A$ than $C$? $AE=7, EC=8$. $AC=15$. So $E$ is roughly in the middle.
The line $EF$ connects these two points.
Since $AF/AB = 7/13$ and $AE/AC = 7/15$, the ratio of segments from $A$ is different.
$7/13 \approx 0.538$ and $7/15 \approx 0.466$.
So $F$ is "lower" on $AB$ than $E$ is on $AC$ relative to the height?
Wait, $AF < AE$ is false. $7 < 7$ is false. $AF=7, AE=7$.
Wait, $AF=7$ and $AE=7$. So $\triangle AFE$ is isosceles with $AF=AE$.
This means $EF$ is perpendicular to the angle bisector of $A$.
Also, the line $EF$ cuts off a triangle $AFE$ similar to $ABC$? No.
In $\triangle AFE$, $AF=7, AE=7$. In $\triangle ABC$, $AB=13, AC=15$.
The ratio $AF/AB = 7/13$ and $AE/AC = 7/15$.
Since $7/13 \neq 7/15$, $EF$ is not parallel to $BC$.
So $EF$ will intersect $BC$ at some point $P$.
Since $AF/AB > AE/AC$ ($7/13 > 7/15$), the point $F$ is "further" from $A$ along the side $AB$ than $E$ is along $AC$ in terms of proportion?
Actually, let's look at the position.
$F$ divides $AB$ in ratio $7:6$. $E$ divides $AC$ in ratio $7:8$.
Let's use Menelaus' Theorem on $\triangle ABC$ and line $P-E-F$? No, $P, E, F$ are collinear.
Wait, $P$ is on $BC$, $E$ is on $AC$, $F$ is on $AB$.
So the line $PEF$ is a transversal cutting the sides of $\triangle ABC$.
Wait, $P$ is on the line containing $BC$. $E$ is on $AC$. $F$ is on $AB$.
Menelaus' Theorem states that for a line intersecting the sides (or their extensions) of a triangle, the product of the ratios of the segments is 1.
Let's apply Menelaus' Theorem to $\triangle ABC$ with transversal line $P-E-F$.
The vertices are $A, B, C$.
The line intersects $AB$ at $F$, $AC$ at $E$, and $BC$ at $P$.
Wait, $F$ is on segment $AB$, $E$ is on segment $AC$.
For the line to intersect the segment $BC$, it would have to pass through the interior of the triangle.
But $F$ and $E$ are on the sides adjacent to $A$.
The segment $EF$ is inside the triangle near $A$.
The line containing $EF$ must intersect the line $BC$ outside the segment $BC$.
Let's check the ratios.
$F$ is between $A$ and $B$. $E$ is between $A$ and $C$.
So the line $EF$ does not intersect the segment $BC$. It intersects the extension of $BC$.
So $P$ is outside the segment $BC$.
Let's determine on which side of $B$ or $C$ the point $P$ lies.
Consider the coordinates or vectors.
Let $A$ be the origin $(0,0)$? No, maybe just use ratios.
Let's use Menelaus' Theorem on $\triangle ABC$ with line $P-E-F$.
The points are $F$ on $AB$, $E$ on $AC$, $P$ on $BC$ (extended).
The theorem states:
$\frac{AF}{FB} \cdot \frac{BP}{PC} \cdot \frac{CE}{EA} = 1$
Wait, the order o