Prove that $B P$ is the bisector of triangle $A B C$, apply the bisector theorem of a triangle and the formula for the square of the bisector of a triangle: $B P^{2}=A B \cdot B C-A P \cdot P C$.

## Solution

In the isosceles triangle $B M N$, point $O$ is the intersection of the altitudes. Therefore, $B P$ is the bisector of triangle $B M N$ and triangle $A B C$. According to the bisector theorem of a triangle, $A B: B C=A P: P C$, so

$$
A B=\frac{A P \cdot B C}{P C}=\frac{5 \cdot 6}{4}=\frac{15}{2}
$$

By the formula for the square of the bisector of a triangle,

$$
B P^{2}=A B \cdot B C-A P \cdot P C=\frac{15}{2} \cdot 6-5 \cdot 4=25
$$

Therefore, $B P=5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8027e864227276d12944g-35.jpg?height=486&width=744&top_left_y=1381&top_left_x=657)

Answer

5.00
$\left[\begin{array}{l}\text { The ratio in which the bisector divides the side. } \\ {[\underline{\text { Properties of bisectors, concurrency }}]}\end{array}\right]$

In triangle $ABC$, points $M$ and $N$ are marked on sides $AB$ and $BC$ respectively, such that $BM = BN$. A line is drawn through point $M$ perpendicular to $BC$, and a line through point $N$ perpendicular to $AB$. These lines intersect at point $O$. The extension of segment $BO$ intersects side $AC$ at point $P$ and divides it into segments $AP = 5$ and $PC = 4$. Find the length of segment $BP$, given that $BC = 6$.