I've basically used this simple lemma for a couple of times. $$\widehat{x}=\widehat{y}\iff PR^2=PS\cdot PQ \iff \triangle PRS \sim \triangle PQR$$[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(6.084218811491537cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -3.396513183785911, xmax = 8.771924439197162, ymin = -0.5995828414010219, ymax = 5.901794569067291; /* image dimensions */ draw(arc((-0.37775248890377966,4.036331597437381),0.472255017709563,-99.05072167718374,-66.36291518836255)--(-0.37775248890377966,4.036331597437381)--cycle, linewidth(2.)); draw(arc((3.94,1.24),0.472255017709563,147.071456358457,179.7592628472782)--(3.94,1.24)--cycle, linewidth(2.)); /* draw figures */ draw((-0.82,1.26)--(3.94,1.24), linewidth(2.)); draw((-0.37775248890377966,4.036331597437381)--(-0.82,1.26), linewidth(2.)); draw((-0.37775248890377966,4.036331597437381)--(3.94,1.24), linewidth(2.)); draw((0.8403908553270368,1.2530235678347603)--(-0.37775248890377966,4.036331597437381), linewidth(2.)); label("$P$",(-1.4445257772530506,1.2264698937426204),SE*labelscalefactor,fontsize(14)); label("$Q$",(3.577119244391969,1.2264698937426204),SE*labelscalefactor,fontsize(14)); label("$R$",(-0.9565289256198356,4.705415190869732),SE*labelscalefactor,fontsize(14)); label("$S$",(0.44449429358520126,1.163502558048012),SE*labelscalefactor,fontsize(14)); label("$x$",(2.742802046438408,1.682983077528531),SE*labelscalefactor,fontsize(14)); label("$y$",(-0.4370484061393163,3.4145848091302606),SE*labelscalefactor,fontsize(14)); /* dots and labels */ dot((-0.82,1.26)); dot((3.94,1.24)); dot((0.8403908553270368,1.2530235678347603)); dot((-0.37775248890377966,4.036331597437381)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Assume that $\widehat{CMD}=\alpha, \widehat{MBC}=\beta ,\widehat{MDC}=\theta$. First we get $\widehat{BCD}=\pi - \theta$. Note that $\widehat{ABC}=\alpha+\beta>\theta\implies \widehat{ABC}+\widehat{BCD}>\pi$, Therefore $AB,CD$ intersect at a point $K$ where $C$ lies between $D,K$. By simple angle chasing in triangles $ABC,ACK$ we obtain $\widehat{AKC}=2\alpha=2\widehat{MBA}$. So $MB,CK$ intersect at a point $L$ where $K$ lies between $C,L$ and also $\widehat{BLK}=\alpha$. $$\left.\begin{matrix} \alpha=\widehat{MCB}=\widehat{MLC}\implies MC^2=MB\cdot ML\\ AM=MC \end{matrix}\right\}\implies AM^2=MB\cdot ML$$ $$\implies \widehat{MBA}=\widehat{MAL}=\alpha=\widehat{MLC}\implies CL^2=CM\cdot CA (\star)$$Also $\alpha=\widehat{ABM}=\widehat{ACB}\implies AB^2=AM\cdot AC \overset{\star}\implies AB=CL$ And $\alpha=\widehat{DMC}=\widehat{MLD}\implies DM^2=DC\cdot DL=DC(DC+CL)=DC(DC+AB)$ We also know that $DM^2=\frac{AD^2+DC^2}{2}-\frac{AC^2}{4}=\frac{AD^2+DC^2}{2}-\frac{AB^2}{2}$ So we finally get $DM^2=DC^2+DC\cdot AB=\frac{AD^2+DC^2}{2}-\frac{AB^2}{2}\implies AB^2+2AB\cdot DC+DC^2=AD^2 \implies AB+DC=AD \blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7.745097287835467cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -3.629767813110425, xmax = 4.115329474725042, ymin = -2.139025087435613, ymax = 3.4925333278391; /* image dimensions */ pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); draw(arc((0.68,0.68),0.4090720398504648,125.90972307917771,163.56020166006078)--(0.68,0.68)--cycle, linewidth(2.)); draw(arc((-0.516619939357202,1.0330877709025847),0.4090720398504648,-113.79430408760756,-76.19285521038485)--(-0.516619939357202,1.0330877709025847)--cycle, linewidth(2.)); draw(arc((2.6371173423608085,0.10251154516096661),0.4090720398504648,163.56020166006078,201.15095358382806)--(2.6371173423608085,0.10251154516096661)--cycle, linewidth(2.)); draw(arc((-1.,3.),0.4090720398504648,-76.19285521038483,-38.54237662950174)--(-1.,3.)--cycle, linewidth(2.)); draw(arc((-0.03323987871440397,-0.9338244581948305),0.4090720398504648,66.15666620873205,103.80714478961515)--(-0.03323987871440397,-0.9338244581948305)--cycle, linewidth(2.)); /* draw figures */ draw((-1.66,-1.56)--(-1.,3.), linewidth(2.)); draw((-1.,3.)--(0.68,0.68), linewidth(2.)); draw((-1.66,-1.56)--(-0.03323987871440397,-0.9338244581948305), linewidth(2.)); draw((-0.03323987871440397,-0.9338244581948305)--(0.68,0.68), linewidth(2.)); draw((-1.66,-1.56)--(2.6371173423608085,0.10251154516096661), linewidth(2.)); draw((-0.516619939357202,1.0330877709025847)--(2.6371173423608085,0.10251154516096661), linewidth(2.)); draw((-0.03323987871440397,-0.9338244581948305)--(-0.516619939357202,1.0330877709025847), linewidth(2.)); draw((-0.35438025100675635,0.030106275123402547)--(-0.19547956706484984,0.06915703758435221), linewidth(2.)); draw((-0.516619939357202,1.0330877709025847)--(-1.,3.), linewidth(2.)); draw((-0.8377603116495542,1.9970185042208173)--(-0.6788596277076476,2.036069266681767), linewidth(2.)); draw((-1.,3.)--(1.434975413244613,-0.36258509448065573), linewidth(2.)); draw((-1.,3.)--(2.6371173423608085,0.10251154516096661), linewidth(2.)); draw((-0.516619939357202,1.0330877709025847)--(-1.66,-1.56), linewidth(2.)); label("$B$",(0.6382171359960913,1.2290013739998693),SE*labelscalefactor,fontsize(14)); label("$A$",(-1.3935073285945507,3.3698117158839613),SE*labelscalefactor,fontsize(14)); label("$M$",(-0.6026347182169853,1.5971662098652863),SE*labelscalefactor,fontsize(14)); label("$D$",(-2.034386857693612,-1.511781292998236),SE*labelscalefactor,fontsize(14)); label("$C$",(-0.24810561701324915,-0.9799876411926335),SE*labelscalefactor,fontsize(14)); label("$K$",(1.2654609304334707,-0.4209225200636667),SE*labelscalefactor,fontsize(14)); label("$L$",(2.560855723293276,-0.011850480213203238),SE*labelscalefactor,fontsize(14)); /* dots and labels */ dot((-1.,3.)); dot((0.68,0.68)); dot((-1.66,-1.56)); dot((-0.03323987871440397,-0.9338244581948305)); dot((-0.516619939357202,1.0330877709025847),uuuuuu); dot((2.6371173423608085,0.10251154516096661)); dot((1.434975413244613,-0.36258509448065573)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $B_1$ be the reflection of $B$ over $AC$ and $B'$ be the point so that $MBCB'$ is a parallelogram. Since $\angle DCB' = \angle MB'C - \angle MDC = \angle MBC - \angle MDC = \angle MCB = \angle CMD$, we have that $DC$ is tangent to $(\triangle MCB').$ Since $MCB'B_1$ is an isosceles trapezoid, $B_1 \in (\triangle MCB').$ The main claim is that $AD$ is tangent to $(\triangle MB'C).$ Then, since $AB$ is the tangent to $(\triangle ABC)$, we'd have that $AB_1$ is tangent to $(\triangle MB'C)$ by symmetry, so that $B_1 \in AD$. This hence implies that $AD = AB_1 + B_1D = AB + DC$, where we used Equal Tangents and symmetry in the last step. Therefore, it suffices only to show that $AD$ is tangent to $(\triangle MB'C).$ To do this, we only need to show that $D$ is the intersection of the tangents to $(\triangle MCB')$ at $B_1$ and $C$. By a well-known property of the symmedian, it would suffice only to show that $MB_1$ is the symmedian in $\triangle MCB_1.$ Observe that $\angle MBA = \angle BCA$ implies that $\triangle ABM \sim \triangle ACB$, so in view of $AC = 2 \cdot AM$ we have that $\sqrt2 \cdot BM = CB \Rightarrow \sqrt2 \cdot B_1M = CB_1.$ Letting $y = MC, x = MB_1$, if we let $P$ be the point where the $M-$symmedian of $\triangle CMB_1$ meets $CB_1$, then we only need to show that $PC = PM.$ Indeed, we have that $PC = \frac{y^2}{y^2 + x^2} \cdot \sqrt2x.$ Therefore, if we let $X$ be the foot from $P$ to $MC$, we have that $CX = PC \cdot \cos {\angle MCB_1} = \sqrt2 x \frac{y^2}{y^2 + x^2} \cdot \frac{y^2 + 2x^2 - x^2}{2 \sqrt2 xy}= \frac{y}{2},$ i.e., $X$ is the midpoint of $MC$. This clearly implies that $CP = PM,$ and so we're done. $\square$