5. In the coordinate plane, an isosceles triangle $\triangle A B C$ is given with vertices $A(2,1), B(6,1)$ and $C(4,5)$. Determine the equations of all lines that are parallel to one of the legs and divide the triangle into two parts of equal area.
First solution.

The line divides the triangle into a trapezoid and an isosceles triangle.

![](https://cdn.mathpix.com/cropped/2024_05_30_5715e7b36c946fa665d7g-36.jpg?height=520&width=574&top_left_y=2153&top_left_x=1255)

Triangles $\triangle A B C$ and $\triangle D B E$ are similar by the AA similarity criterion with a similarity ratio $k$, and the ratio of their areas is $k^{2}$. $1$ POINT

The area of triangle $\triangle A B C$ is twice the area of triangle $\triangle D B E$, so $k^{2}=2$, and $k=\sqrt{2}$. $1$ POINT From $|A B|=4$ and $k=\sqrt{2}$, it follows that

$\frac{|A B|}{|D B|}=\sqrt{2}$, so $|D B|=\frac{|A B|}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2 \sqrt{2}$.

Thus, the coordinates of point $D$ are $D(6-2 \sqrt{2}, 1)$.

The equation of line $A C$ is obtained by solving the system:

$A(2,1), C(4,5) ; \quad y=a x+b$

$1=2 a+b$

$\underline{5=4 a+b}$

$y=2 x-3$

(Remark. To get these 2 POINTS, it is sufficient to determine that $a=2$.)

The desired line is parallel to line $A C$ and has the same slope, passing through point $D$.

$D(6-2 \sqrt{2}, 1), a=2 ; \quad y=a x+b$

$1=2(6-2 \sqrt{2})+b$

$b=-11+4 \sqrt{2}$

$y=2 x-11+4 \sqrt{2}$

$1$ POINT

Similarly, the line parallel to side $B C$ passing through the point $(2+2 \sqrt{2}, 1)$ is obtained.

The equation of line $B C$ is obtained by solving the system:

$B(6,1), C(4,5) ; \quad y=a x+b$

$1=6 a+b$

$\underline{5=4 a+b}$

$y=-2 x+13$

![](https://cdn.mathpix.com/cropped/2024_05_30_5715e7b36c946fa665d7g-37.jpg?height=546&width=557&top_left_y=1509&top_left_x=1275)

2 POINTS

(Remark. To get these 2 POINTS, it is sufficient to determine that $a=-2$. This follows from the symmetry with the previous case, so this conclusion can be made without recalculating $a$.)

The desired line is parallel to line $A C$ and has the same slope, passing through the point $(2+2 \sqrt{2}, 1)$.

$(2+2 \sqrt{2}, 1), a=-2 ; \quad y=a x+b$

$1=-2(2+2 \sqrt{2})+b$

$b=5+4 \sqrt{2}$

$y=-2 x+5+4 \sqrt{2}$

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