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Original Drawing

Personal Attempt At Creating Congruent Triangles

$\overline{AC}$ is a radius of the large circle, and a diameter of the small circle. A line through $A$ cuts the small circle at $X$ and the large circle at $B$. Show that $X$ is the midpoint of $\overline{AB}$.

My attempt:

Since they are both radii, $AC=AB$. Then, drawing a line between $X$ and $C$, we have created another side that is shared between the two triangles: $\triangle ACX$ and $\triangle BCX$. If we can show that angle $ACX$ is equal to angle $BCX$ (this is where I'm stuck), then using SAS $\triangle ACX$ is congruent to $\triangle BCX$. Therefore, $AX=BX$ and $X$ is the midpoint of $\overline{AB}$.

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$AXC$ is a right triangle (in fact is inscribed into the small circle and $AC$ is a diameter) and it is congruent to $BXC$ triangle (it is right triangle as well, $AC=R=BC$ and $XC$ is in common ), so $AX=XB$.

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