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Let each apex (point) of a triangle be apex $A$, apex $B$, and apex $C$. Let each length then be $\overline {AB},\overline {AC},\overline {BC}$.

If triangle is cut in half, through an imaginary line from apex $A$ down to the exact middle (bisection) of $BC$, then will the new angle at apex $A'$ (angle of apex $C$ to apex $A$ to the point at $\frac{\overline {BC}}2$) be exactly equal to half the old angle at apex $A$ (that is equal to angle $\frac {B\hat AC}2$)?

That is, does new angle $A'=\frac A2$ ?

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    $\begingroup$ The center of mass of a triangle is the intersection of the three medians (en.wikipedia.org/wiki/Centroid). The incenter of a triangle (center of the inscribed circle) is the intersection of the three angle bisectors. They only coincide for equilateral triangles (math.ucr.edu/~res/inprogress/concurrence2.pdf) $\endgroup$ – lulu Sep 26 '15 at 13:40
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    $\begingroup$ In other words: NO, unless $AB=AC$. $\endgroup$ – Aretino Sep 26 '15 at 13:45
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Ratio of sides and cut parts of opposite side is same for angle bisection at A. $ AB/AC = AM/MC $ So, unless the triangle is isosceles ( AB= AC) it will not be so.

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