Let $AB \cap CD=X;DX\cap AM=H,BI\perp DX(I\in DX);AC\cap BD=Y$ $\therefore AM\perp DX$ $\therefore$ To prove $\angle ADB=\angle CAM \Leftrightarrow \angle ABY=\angle ACX\Leftrightarrow X,B,Y,C$ are concyclic$\Leftrightarrow AB\cdot AX=AY\cdot AC$ Let $AB=\lambda AX;AB=a;\angle AXD=\theta$ $\therefore IH=CH=\lambda acos\theta\Rightarrow XC=(1+\lambda)acos\theta\Rightarrow CD=\frac{a}{cos\theta}-(1+\lambda)acos\theta$ and$BX=(1-\lambda)a\Rightarrow AB\cdot AX=\lambda a^2$ We apply The Menelaus’ to $\triangle AXC$ and line $BYD$ $\therefore \frac{AB}{BX}\cdot \frac{XD}{DC}\cdot \frac{CY}{YA}=1\Rightarrow \frac{\lambda a}{(1-\lambda)a}\cdot \frac{\frac{a}{cos\theta}}{\frac{a}{cos\theta}-(1+\lambda)acos\theta}\cdot \frac{CY}{YA}=1$ We can calculate $\frac{CY}{YA}=\frac{1-\lambda -(1-\lambda^2)cos^2\theta}{\lambda}$ And $AC^2=AH^2+CH^2=a^2[1-(1-\lambda^2)cos^2\theta]$ $\therefore AY\cdot AC=\frac{1}{1+\frac{CY}{YA}}\cdot AC^2=\lambda a^2$ $\therefore AB\cdot AX=AY\cdot AC$ So we are done.

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