HEEEEEEEEEEELLP

Question

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. 

Answer 1

I hope you drew a picture! It would be pretty hard to figure out all the parts of this problem without looking at it, so please draw rectangle ABCD right now.

Given: In rectangle ABCD, the angle bisector of ∠A intersects side DC  at point M

Notice that ∠MAD must be half of 90, so ΔMAD is a 45-45-90 triangle, meaning that AD=MD and AM=DM√2

Given: The angle bisector of ∠C intersects side AB at point N.

Same deal here, because ∠NCB must be half of 90, ΔNCB is a 45-45-90 triangle,so BC=BN and CN=BC√2.

Additionally, ABCD is a rectangle, meaning AD=BC. By the transitive property of equality, I can use the equations AD=MD, AD=BC, and BC=BN to say that AD=BN. Using substitution I can also say that CN=AM. I can also say that CD=AB, and then using the segment addition postulate and subtraction I can say that MC=NA.

Given: The length CM is equal to the length AM and it is 6 in longer than the length DM

If AM = DM+6 then that means AM-6=DM, and we already said that AM=DM√2. Using substitution we can say that AM=(AM-6)√2

We can do some algebra here to solve for AM

AM=(AM-6)√2

AM=AM√2-6√2

AM-AM√2=-6√2

AM(1-√2)=-6√2

AM=(-6√2)/(1-√2)

I have to simplify rational expressions, so I'm gonna move the radical here (with the help of a conjugate), but your calculator could help you round if you don't recognize this.

AM=(-6√2)/(1-√2)

AM=(-6√2)(1+√2)/(1-√2)(1+√2)

AM=(-6√2-12)/(1-√2+√2-2)

AM=(-6√2-12)/(-1)

AM=6√2+12

Because CM=AM, and we already said CM=BN and that MC=NA, I see that AM=CM=BN=AN, so ANCM is a rhombus. AM=6√2+12, so the perimeter of ANCM is 4×(6√2+12), or 24√2+48



Let me know if you need clarification on any of these steps, or if I made a mistake!