A geometry problem by Chung Kevin

As $M$ is the midpoint of $AB$ , we have that $|AM| = |BM| = |MC| = |MD|$ . So $M$ is the center of a circle on which $A, B, C$ and $D$ lie. Then as both $\angle DAB$ and $\angle DCB$ are subtended by the same chord $DB$ , we have that $\angle DCB = \angle DAB = 40^{\circ}$ .

Finally, as $\Delta ABC$ is right-angled at $C$ we see that $\angle DCA = 90^{\circ} - \angle DCB = \boxed{50^{\circ}}$ .

if you imagine that this shape is a shape with two headed arrows D and B. When it is folded the line AC is created. This means that angle DAB=Angle DCB= 40 so angle DCA= 90-40=50.

Put am=MB=L, CB=N and Cm=y. cos (b)=(N÷(2.L))={N^2+L^2-y^2}÷{2.N.L}. so L^2=y^2 so L=y ,,,,,,@∆ AMD , am=MD so. Angle (M D A) =40° and <AMD =100°. @∆ MCA cm=ma ,put Angle Mac=MCA=t and <CMA=u so U=180-2.t ,so Angle CMD=100+180-2.t=280-2.t ,,@∆ CMD Angle MCD= MDC =(180-(280-2.t))÷(2)=t-50°. so required Angle (ACD)=k=t-(t-50)=t-t+50 so k=50°########

As $\bigtriangleup ABC$ is right triangle, we have $\overline{MA} \cong \overline{MB} \cong \overline{MC} \cong \overline{MD}$ , therefore quadrilateral $ABDC$ is cyclic, hence $\angle ADB=90^{\circ}$ and $\angle DCA = \angle ABD = 50^{\circ}$

Know AM = MB = MC = MD. Triangles CDM, CMB, AMD, and CMA are all isosceles.
Angles ADC + ACD + CAD = 180.
Angle ADC = 40 - CDM. Angle ACD = ACM - DCM. Angle CAD = CAM + 40. Substitute into second line. (40 - CDM) + (ACM - DCM) + (CAM + 40) = 180. 80 - CDM - DCM + ACM + CAM = 180. Since CDM = DCM and CAM = ACM, 2(ACM) - 2(DCM) = 100.
2(ACM - DCM) = 100. ACM - DCM = 50.