In rectangle abcd, the angle bisector of ∠a intersects side dc at point m and the angle bisector of ∠c intersects side ab at point n. the length of cm is equal to the length of am and it is 6 in longer than the length of dm . find the perimeter of quadrilateral ancm. plz

AM is an angle A bisector and as ABCD is rectangular, m∠A=90^{0}, then m∠DAM=45^{0}.
Consider right triangle ADM. m∠D=90^{0} (because ABCD is rectangle), so m∠DMA=180^{0}-90^{0}-45^{0}=45^{0}. Therefore, ΔADM is isoscales and that's why AD=DM. Let x be the length of DM. Since ΔMC is 6 in longer than DM, MC has length x+6. The length of AM is equal to the length of CM, so AM=x+6.
Return to the triangle ADM, by the Pythagorean theorem AM^2=AD^2+DM^2.
(x+6)^2=x^2+x^2
x^2+12x+36=2x^2
x^2-12x-36=0
D=(-12)^2-4*(-36)=144+144=288


- incorect answer, because x_{2}<0.
The perimeter of quadrilateral ANCM is equal to 4(x+6)=.

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