ABCD is a rhombus. M and N are points on BD such that DN = MB. Prove that Triangle DNC is congruent to triangle BMC.

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It is well-known that in any parallelogram the opposite angles are equal. A rhombus is a parallelogram, so the angles `ABC` and `ADC` are congruent.

Also, it is known that any diagonal of a rhombus bisects the corresponding angles. Therefore angles `ADN` and `CDN` are congruent, and `ABM` and `CBM` are congruent. From this and the first paragraph we infer that the angles `CDN` and `CBM` are congruent (both are halves of the congruent angles).

Now we can finish the proof. The triangles `DNC` and `BMC` have two pairs of congruent sides: ` ``DN = BM` by the conditions and `DC = BC` by definition of rhombus. Also the angles between these sides are equal in both triangles, `CDN = CBM` as proved above. Therefore these triangles are congruent by the side-angle-side rule (SAS).

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