I drew it to scale because I was trying to notice something, but I cannot figure anything out.
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Then this becomes the case of a triangle in a semi-circle. – peterwhy Sep 24 '15 at 23:08
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I noticed that, but I'm afraid to make observations due to this because it isn't given that CF becomes the diameter. – Charles Sep 24 '15 at 23:18
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1"If": let $E$ be the mid-point of $CF$, then $CE = AE = EF$. Then $A,C,F$ is on a circle with centre $E$. And since $CEF$ is a straight line, it is also a diameter. – peterwhy Sep 24 '15 at 23:21
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That makes complete sense. And since arc GF (longways) is 180 degrees, angle A must be a right angle? – Charles Sep 24 '15 at 23:30
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"If": In $\triangle ABC$, let $E$ be the mid-point of $BC$, then $BE = AE = EC$. Then $A,B,C$ is on a circle with centre $E$.
And since $BEC$ is a straight line, it is also a diameter. $\angle A = 90^\circ$ because it is an angle in semi-circle.
"Only if": Duplicate $\triangle A'CB \cong\triangle ABC$, with $A'$ and $A$ on different sides of line $BC$. Then $ABA'C$ is a rectangle because $\angle A = 90^\circ$.
The median from $A$ to the mid-point of $BC$ is half of the diagonal $AA'$ of the rectangle. In a rectangle, diagonals are of equal length and bisect each other.
peterwhy
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An alternative. By the law of cosines, $$ 4m_a^2 = 2b^2+2c^2-a^2,\tag{1}$$ hence $4m_a^2=a^2$ implies: $$ b^2+c^2=a^2\tag{2}$$ from which $\widehat{BAC}=\frac{\pi}{2}$.
Jack D'Aurizio
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