If: Let $P$ be the point on $BC$ such that $PA = PB$. Then $AC = 2DM = \frac{2\cdot CP \cdot MB}{PB} = \frac{CP\cdot AB}{BP}$, so $\frac{AC}{AP} = \frac{AB}{BP}$. Then by converse of the Angle Bisector Theorem, $\angle CAP = \angle BAP$, so $\angle A = 2\angle B$. Only If: Let $P$ be the point on $BC$ such that $PA = PB$. Then $\angle PAB = \angle PBA$, so $\angle CBA = \angle CAP$, and triangles $\triangle CAP$ and $\triangle CBA$ are similar. Then $\frac{AC}{AB} = \frac{CP}{PA}=\frac{CP}{PB}=\frac{DM}{MB}=\frac{2DM}{AB}$, so $AC=2MD$.

Nice idea, nsun48! The solution could be even faster, seeing that $MP\parallel CD$. Then $\hat A=2\hat B\implies\frac{AC}{AB}=\frac{CP}{BP}=\frac{DM}{MB}$; as $AB=2BM$, we get $AC=2DM$. Conversely, $AC=2DM\implies AP$ angle bisector of $\angle BAC$ and, with $AP=PB\iff \hat B=\widehat{BAP}$, we are done! Congratulations! My idea was based on the relation $BC^2=AC(AB+AC)\iff \hat A=2\hat B$, then working it with Carnot lemma. Best regards, sunken rock

This is my solution for this problem.

Denote by $M'D'$ the image of $MD$ in homothety $\mathcal{H}(H,2)$. It is well-known that $M'D'$ is a chord on the circumcircle $\omega$ of $\triangle ABC$ and we also have $M'D' = 2MD$. It suffices to note that the inscribed angles corresponding to $AC$ and $M'D'$ are $\angle B$ and $|\angle A - \angle B|$, respectively. Done.

Dear all

good job MahanBabol. and how about the other side?

just put $\measuredangle A=\measuredangle (B+\alpha )$ after some calculating you'll get $cos\measuredangle A=\frac{sin\measuredangle A.cos\measuredangle B }{sin \measuredangle B } - 1 \equiv 1-sin\measuredangle B.sin\measuredangle \alpha=\frac{cos^{2}\measuredangle B.sin\alpha }{sin\measuredangle B} \equiv \frac{1}{sin\alpha }=\frac{cos^{2}\measuredangle B + sin^{2}\measuredangle B }{sin\measuredangle B }$ which means : $\alpha = \measuredangle B$ And this is the other side Mr. Down With Israel.

We can use angle chasing: Denote $N$ is the midpoint of $AC$, so we have $(DN,AB) \equiv (AB,AN) (mod \pi)$ \[(AB,AC) \equiv 2(BC,BA) \Leftrightarrow (DN,AB) \equiv 2(MN,AB) (mod \pi)\] \[\Leftrightarrow (DN,MN) + (MN,AB) \equiv (MN,AB) (mod \pi)\] \[\Leftrightarrow(DN,MN) \equiv (MN,DM) (mod \pi) \Leftrightarrow DM=DN \Leftrightarrow DM = \dfrac{1}{2}AC\]

nsun48 wrote: In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$. Prove that $\angle A = 2\angle B$ if and only if $AC = 2 MD$. Let $N$ is the midpoint of side $AC$ Thus triangle $DMN$ is isosceles at D Thus It easily follows that to prove

duanKHTN wrote: nsun48 wrote: In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$. Prove that $\angle A = 2\angle B$ if and only if $AC = 2 MD$. Let $N$ is the midpoint of side $AC$ Thus triangle $DMN$ is isosceles at D Thus It easily follows that to prove duanKHTN wrote: nsun48 wrote: In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$. Prove that $\angle A = 2\angle B$ if and only if $AC = 2 MD$. Let $N$ is the midpoint of side $AC$ Thus triangle $DMN$ is isosceles at D Thus It easily follows that to prove nsun48 wrote: In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$. Prove that $\angle A = 2\angle B$ if and only if $AC = 2 MD$. Let $N$ is the midpoint of side $AC$ Thus triangle $DMN$ is isosceles at D Thus It easily follows that to prove

Let $D'$ is the midpoint of $BC$,so $MD'||AC$ and $AC=2MD'$,now let $MD'=MD$ and ${\angle}A=2X$ ($X$ is real),so ${\angle}M=2X$,hence ${\angle}MDD'=X$,but $DD'=BD'$,(angle $D$ is right and $D'$ is the midpoint),so ${\angle}ABC=X$,hence we are done.

Let $\angle CBA=\theta\implies\angle CAB=2\theta$. Extend $BA$ to a point $K$ such that $AC=AK$. So, $\angle CKA=\theta$. So, $\triangle CBD\cong\triangle CKD$. So, $BD=\frac{BK}{2}=\frac{b+c}{2}$. So, $DM=\frac{b}{2}$ and for the converse, extend $BD$ to a point $N$ such that $BD=DN$. We know $DM=\frac{b}{2}$ and $BM=\frac{c}{2}\implies DN=\frac{b+c}{2}$. So, $AN=(b+c)-c=b=AC\implies \angle NCA=\theta\implies \angle CAB=2\theta$.

It is well known that, $\angle A = 2\angle B \Leftrightarrow {a^2} - {b^2} = bc \Leftrightarrow 2c(MD) = bc.$ Done!