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

First, let's find 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 $AE = AF = x$, $BF = BD = y$, $CD = CE = z$.
We have:
$x + y = c = 13$
$y + z = a = 14$
$z + x = b = 15$
Adding these equations: $2(x+y+z) = 13+14+15 = 42$, so $x+y+z = 21 = s$.
Thus:
$x = s - a = 21 - 14 = 7$. So $AE = AF = 7$.
$y = s - b = 21 - 15 = 6$. So $BF = BD = 6$.
$z = s - c = 21 - 13 = 8$. So $CD = CE = 8$.

Let's check: $AE+EC = 7+8=15=b$, $AF+FB = 7+6=13=c$, $BD+DC = 6+8=14=a$. Correct.
The points are $F$ on $AB$ and $E$ on $AC$.
$AF = 7$, $FB = 6$.
$AE = 7$, $EC = 8$.

We are considering the line $EF$. We want to find the intersection $P$ of line $EF$ with line $BC$.
Let's set up a coordinate system or use Menelaus' theorem.
Let's use Menelaus' theorem on $\triangle ABC$ with transversal line $EFP$.
Wait, $E$ is on $AC$, $F$ is on $AB$, and $P$ is on $BC$.
The points $E, F, P$ are collinear.
According to Menelaus' theorem applied to $\triangle ABC$ and line $EFP$:
$\frac{AF}{FB} \cdot \frac{BP}{PC} \cdot \frac{CE}{EA} = 1$.
Note the order of points. The vertices are $A, B, C$.
The line intersects $AB$ at $F$, $BC$ at $P$, $AC$ at $E$.
The segments are directed.
Let's use lengths.
$F$ is on segment $AB$. $E$ is on segment $AC$.
$P$ is on line $BC$.
The ratio is $\frac{AF}{FB} \cdot \frac{BP}{PC} \cdot \frac{CE}{EA} = 1$.
Let's verify the direction.
$A \to F \to B$. $F$ is between $A$ and $B$. So $AF/FB$ is positive.
$B \to P \to C$. $P$ is on the extension of $BC$ or on the segment?
$E$ and $F$ are on the sides $AC$ and $AB$. The line $EF$ connects points on two sides.
Usually, the line connecting points on two sides intersects the third side externally.
So $P$ should be outside the segment $BC$.
Let's check the ratios.
$AF = 7$, $FB = 6$. So $AF/FB = 7/6$.
$CE = 8$, $EA = 7$. So $CE/EA = 8/7$.
The equation is $\frac{7}{6} \cdot \frac{BP}{PC} \cdot \frac{8}{7} = 1$.
$\frac{7}{6} \cdot \frac{8}{7} = \frac{8}{6} = \frac{4}{3}$.
So $\frac{4}{3} \cdot \frac{BP}{PC} = 1$.
$\frac{BP}{PC} = \frac{3}{4}$.
Here $BP$ and $PC$ are lengths of segments on the line $BC$.
Since $P$ is outside $BC$, $P, B, C$ are collinear.
Let's determine the position of $P$.
The line $EF$ intersects $AB$ and $AC$.
Let's visualize. $A$ is the top vertex. $B$ is bottom left, $C$ is bottom right.
$F$ is on $AB$ closer to $B$ (since $AF=7, FB=6$, wait $AB=13$, $AF=7$ is slightly more than half).
$E$ is on $AC$ closer to $A$ (since $AE=7, EC=8$, $AC=15$).
Wait, $AF=7$ and $AE=7$. So $AF=AE$.
This means $\triangle AFE$ is isosceles with $AF=AE$.
So $\angle AFE = \angle AEF$.
Also $\angle AFE = \angle BFP$ (vertically opposite? No, $F$ is on $AB$, $P$ is on $BC$).
Let's check the position of $P$.
The transversal $EFP$ cuts $AB$ at $F$ and $AC$ at $E$.
By Menelaus, $P$ divides $BC$ externally in ratio $3:4$.
Let's denote coordinates or vectors to be sure about the order $B, C, P$ or $P, B, C$ or $B, P, C$.
The ratio $\frac{BP}{PC} = \frac{3}{4}$.
Since $P$ is on the line $BC$, and $E, F$ are on the segments $AC, AB$, $P$ must be outside the segment $BC$.
Let's check the Menelaus configuration again.
Vertices $A, B, C$. Line intersects $AB$ at $F$, $AC$ at $E$, $BC$ at $P$.
The product of ratios is 1.
$\frac{AF}{FB} \times \frac{BP}{PC} \times \frac{CE}{EA} = 1$.
$AF=7, FB=6 \implies AF/FB = 7/6$.
$CE=8, EA=7 \implies CE/EA = 8/7$.
Product is $(7/6) \times (BP/PC) \times (8/7) = 1$.
$(4/3) \times (BP/PC) = 1 \implies BP/PC