Problem 1041

a) In this case, the lines drawn pass through the midpoints of the sides of the triangle.

b) Let point $M_{1}$ be located on side $A B$ of triangle $A B C$ and be different from the midpoint of this side; $M_{2}$ - a point on side $B C$, such that $M_{1} M_{2} \| A C ; M_{3}$ - a point on side $A C$, such that $M_{2} M_{3} \| A B$ and so on.

If $A M_{1}: M_{1} B=x: y$, then by the theorem of proportional segments $B M_{2}: M_{2} C=y: x, C M_{3}: M_{3} A=x: y$, $A M_{4}: M_{4} B=y: x, B M_{5}: M_{5} C=x: y$ $C M_{6}: M_{6} A=y: x, A M_{7}: M_{7} C=x: y=A M_{1}: M_{1} B$. Therefore, point $M_{7}$ coincides with point $M_{1}$.

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Two circles intersect at points $A$ and $B$. A line through point $A$ intersects the circles at points $C$ and $D$, and a line through point $B$ intersects the circles at points $E$ and $F$ (points $C$ and $E$ are on one circle, $D$ and $F$ are on the other). Prove that $\angle C B D=\angle E A F$.

## Hint

Apply the theorem of inscribed angles.

## Solution

Consider the case when lines $C D$ and $E F$ are either parallel or intersect outside the given circles. Since

$$\angle C B E=\angle C A E, \angle D B F=\angle D A F$$

then

$$\angle C B D=180^{\circ}-(\angle C B E+\angle D B F)=180^{\circ}-(\angle C A E+\angle D A F)=\angle E A F .$$

Similarly for the other cases.

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