# Task 10.5

The altitudes AD and $\mathrm{BE}$ of an acute-angled triangle $\mathrm{ABC}$ intersect at point $\mathrm{H}$.

The circumcircle of triangle $\mathrm{ABH}$ intersects sides $\mathrm{AC}$ and $\mathrm{BC}$ at points $\mathrm{F}$ and $\mathrm{G}$, respectively. Find FG, if $\mathrm{DE}=5$ cm.

## Number of points 7

#
# Answer:

$\mathrm{FG}=10 \mathrm{~cm}$

## Solution

We have only one numerical given.

So, the length of FG will either be equal to ED or a multiple of it. It doesn't seem to be equal. It should be larger (judging by the diagram), by some factor. What do we know about such relationships? There is the midline of a triangle, which is half the length of the base. Let's try to show that ED is the midline in triangle FCG. We need E to be the midpoint of FC. Let's look at triangle FCB. It has a height BE. Let's try to show that it is also a bisector (since we also have a circumscribed circle, and a circle involves angles). On FH, there are two angles: FAH and HBF. They are equal. Let's denote them as $\alpha$. Then from the right triangles, we get $\angle \mathrm{C}=90^{0}-\alpha$ and $\angle \mathrm{CBE}=90^{0}-\left(90^{0}-\alpha\right)=\alpha$. We have confirmed that BE is both a height and a bisector, and therefore a median. Similarly, $\mathrm{AD}$ is a median for triangle CAG. Thus, ED is the midline in triangle FCG, and therefore $\mathrm{FG}=10 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ad017f5a8e5e3e848308g-3.jpg?height=430&width=515&top_left_y=633&top_left_x=1367)