What is a simpler method to find the perimeter of triangle AMN?

[veronica1999]

Triangle ABC has side-lengths AB=12,BC=24,and AC=18. The line
through the incenter of ABC parallel to BC intersects AB at M and AC at
N. What is the perimeter of triangle AMN?
(A)27 (B)30 (C)33 (D)36 (E)42

My friend is saying my solution is too messy and there was no need for the heron's formula. (without telling me how he solved it :()
Can someone show me a simpler way to solve this?

I used the herons formula to get the area of ABC.
Area is 135.
Then i found the radius of the incircle.
6r+9r+12r = 135
r= 5

The height of the triangle ABC is 45/5 ,
so using the rate of 45/5 : 25/5 which is 9:5
I got the lengths of all the other sides.

12: X = 9:5
60/9

24: X = 9:5
120/9

18 : X = 9:5

90/9

270/9 = 30

 

[caffeinemachine]

Triangle ABC has side-lengths AB=12,BC=24,and AC=18. The line
through the incenter of ABC parallel to BC intersects AB at M and AC at
N. What is the perimeter of triangle AMN?
(A)27 (B)30 (C)33 (D)36 (E)42

My friend is saying my solution is too messy and there was no need for the heron's formula. (without telling me how he solved it :()
Can someone show me a simpler way to solve this?

I used the herons formula to get the area of ABC.
Area is 135.
Then i found the radius of the incircle.
6r+9r+12r = 135
r= 5

The height of the triangle ABC is 45/5 ,
so using the rate of 45/5 : 25/5 which is 9:5
I got the lengths of all the other sides.

12: X = 9:5
60/9

24: X = 9:5
120/9

18 : X = 9:5

90/9

270/9 = 30

Let $I$ be the incenter.
Let $AI$ meet $BC$ at $D$.

So $\frac{AB}{BD}=\frac{AC}{CD}$.

This gives the exact lengths of $BD$ and $DC$.

Note that $CI$ is the angle bisector of angle $ ACD$.

Again $\frac{AI}{ID}=\frac{AC}{CD}$.

So now you know the value of $\frac{AI}{ID}$.

Note that using similarity in $\Delta ADC$ we have $\frac{AN}{NC}=\frac{AI}{ID}$.

Can you finish?