Let $I$ be the incenter and $I_A$ be the excenter. The angle condition implies $MA$ is tangent to $(BXA)$, so $MA^2 = MX \cdot MB$. On the other hand, since $(A, D; I, I_A) = -1$ we have $MA^2 = MI \cdot MI_A$. Hence $BXII_A$ is cyclic. But it is well-known that $C$ also lies on this circle. Now $\angle AXC = \angle A/2 + (180^{\circ} - \angle BXC) = \angle A /2 + (180^{\circ} - \angle BIC) = 90^{\circ}$.

Solution. Let $I$ and $I_A$ be, as usual, the incenter and $A$-excenter of $\bigtriangleup ABC$. Since $(A,D;I,I_A)=-1$ and taking into account the angle condition, we get $$MX\cdot MB=MA^2=MI\cdot MI_A$$i.e. $IXBI_A$ is cyclic, thus $$\angle AXC=\angle MXA+\angle MXC=\angle BAD+180^\circ-\angle BIC=\angle BAD+90^\circ-\angle DAC=90^\circ$$as required. $\blacksquare$

Essentially the same problem as 2014 Taiwan TST Round 1 P4.

Let $AH\perp BC$ at $H$. The condition implies $\angle MXD=\angle MDB$. Then we can show that $X,M,D,H$ are concyclic by angle chasing. Thus $A,X,H,C$ are concyclic, and $AX\perp XC$.

If we draw a perpendicular from A to BC(point P), the result is that's enough to prove AXPC is cyclic. $ \Rightarrow $ It's enough to prove $ \angle PXB = \angle ADB \Rightarrow $ It's enough to prove XMDP is cyclic $ \Rightarrow $ It's enough to prove $ \angle DXM = \angle DPM $. We khow that AM=MD=PM $ \Rightarrow $ $ \angle ADB = \angle DAC + \angle ACB = \angle DPM $. $ \Rightarrow $ It's enough to prove $ \angle DXM = \angle DAC + \angle ACB $ . Claim1: $ \angle ABM = \angle MAX $ . Proof : $ \angle AMB = \angle AMB $ and $ \angle MXA = \angle MAB \Rightarrow MXA \sim MAB $ . Claim2: $ \angle MBD = \angle MDX $ . Proof : By claim1 $ AM^2 = MX.MB $ and AM=MD $ \Rightarrow MD^2 = MX.MB \Rightarrow \angle MBD = \angle MDX $ . $ \Rightarrow \angle ABD = \angle XAM + \angle XDM $ , and the sum of the interior angles of the triangle AXD and ABD complete our proof.

Attachments:

Here's a nice sol. Let $U$ be the meeting of $(ABX)$ with $AC$ and $Y=BX\cap AC$. By POP and angle chase, $AU=AB$ and $\angle UXA=\angle AXY$. Hence, as $XA$ bissects $UXY$, we just need $(U,Y;A,C)=-1$. But everything is now chaseable: $AU/AY=CU/CY \iff AB/AB+AC=AY/CY$. But $AY/CY$, by sine law, is just $AB/BC . sin \angle ABY/sin \angle YBC$. But $sin \angle ABY/sin \angle YBC=sin ABM/sin MBD=BD/AB$, again by sine law on $ABM,MBD$ using $MA=MD$. Thus, substituting everything we just need $AB/AB+AC=BD/BC \iff 1 + AC/AB=1+CD/DB=BC/BD$, which is true by multiplying by $BD$. Done =)

The condition implies $\triangle MAX \sim \triangle MBA$, so $MX\cdot MB=MA^2=MD^2$. Therefore $\overline{AMD}$ is tangent to $(BDX)$. Let $E$ be the foot of the $A$-altitude. Using $ME=MA=MD$ and the tangency, we have $\measuredangle MED=\measuredangle EDM=\measuredangle BXD=\measuredangle MXD$, so $DEMX$ is cyclic. Then $\measuredangle CEX=\measuredangle DEX=\measuredangle DMX=\measuredangle MAX+\measuredangle AXM=\measuredangle CAM+\measuredangle MAX=\measuredangle CAX$, hence $ACEX$ is cyclic so $\angle AXC=90^\circ. $\blacksquare$