Help me find a side of triangle given 2 sides and one circle

Question

In $\triangle ABC$, $BC=4$, $AB=2\sqrt{19}$. The center of a circle, which crosses the midpoints of the sides of $\triangle ABC$, lies on the bisector of $\angle ACB$. Find the lenght of $AC$.

Answer 1

The circle described is the Nine-point circle of the triangle, so it also crosses B', the foot of B on AC. We may also call M:the midpoint of BC, N:the midpoint of AC, O:the centre of the Nine-point circle.

$_\triangle$MCO and $_\triangle$NCO have (OM=ON=radius,CO=CO=common,$\angle$MCO=$\angle$NCO)

$_\triangle$MCO and $_\triangle$B'CO have (OM=OB'=radius,CO=CO=common,$\angle$MCO=$\angle$B'CO)

Both are SSA ambiguous forms (that is, they have two same Sides and one same Angle but not the one between the Sides).

However there can be at most two triangles with SSA form, so $_\triangle$MCO is congruent to one of the other two ($_\triangle$NCO can`t be congruent to $_\triangle$B'CO or else altitube would also be median and the triangle would have BC=BA).

So there are two options:

• CM=CN=2 which means that AC=4. This can`t be the case because it violates triangular inequality.
• CM=CB'=2. Now Pythagorean theorem gives BB'=$\sqrt{12}$ and B'A=8. So in this case AC=10

Sorry for poor English and bad edit, first post.

edits: correcting speling and discard of first option