geometry question involving similar triangles

[cmfriedland]

I've been given the following assignment:

There is a right triangle ABC with the right angle at A. The base AB = 4, the side AC = 12.
On the base AB there is a point M. There is a line, parallel to AC, connecting M to point N on the hypotenuse. There is another line, parallel to AB, connecting point N to point P on side AC. In other words, AMNP is a rectangle.

The question has two parts;
First, calculate the length of AP, assuming that the length of AM is 1. That seems pretty easy. Because NP and AB are parallel, we can use the intercept theorem to figure out the dimensions of the smaller triangle. We already know that NP = AM because they are opposite sides of a triangle, i.e. NP = 1. We know that AB is 4. So, 1/4 = CP/12, i.e. CP = 3. Then you subtract 3 from 12 to find out that AP is 9. So far, so good, right?

It's the second part I can't figure out. Here, they say to assume that the location of point M on AB is not given... so the length of AM is simply x. The assignment asks us to prove that AP = 12 - 3x, regardless of the value of x. Obviously AP = 12 - 3x when x = 1 as in the first part, but what theorem (or what else?) could I use to prove that it is true no matter what the value of x? Thanks!

 

[soroban]

Hello, cmfriedland!

There is a right triangle ABC with the right angle at A. The base AB = 4, the side AC = 12.
On the base AB there is a point M. There is a line, parallel to AC, connecting M to point N on the hypotenuse.
There is another line, parallel to AB, connecting point N to point P on side AC.
In other words, AMNP is a rectangle.

The question has two parts;
First, calculate the length of AP, assuming that the length of AM is 1.

That seems pretty easy. Because NP and AB are parallel, we can use the intercept theorem to figure out the dimensions of the smaller triangle.
We already know that NP = AM because they are opposite sides of a rectangle , i.e. NP = 1.
We know that AB is 4. So, 1/4 = CP/12, i.e. CP = 3. Then you subtract 3 from 12 to find out that AP is 9.
So far, so good, right? . Right!

It's the second part I can't figure out.
Here, they say to assume that the location of point M on AB is not given.
So the length of AM is simply x.
The assignment asks us to prove that AP = 12 - 3x, regardless of the value of x.

  - C o
  :   | *
  :   |   *
  :   |     *   N
  : P o - - - o
 12:   |       | *
  :   |       |   *
  :  y|       |y    *
  :   |       |       *
  :   |       |         *
      o - - - o - - - - - o
      A   x   M    4-x    B
      : - - - - 4 - - - - :

\(\displaystyle \text{Let }y \,=\,AP \,=\,MN.\)

\(\displaystyle \text{Since }AB = 4\text{ and } AM = x,\,\text{ then: }\:MB \,=\,4-x\)

\(\displaystyle \Delta NMB \sim \Delta CAB \quad\Rightarrow\quad \frac{y}{4-x} \:=\:\frac{12}{4} \quad\Rightarrow\quad y \:=\:12-3x\)

\(\displaystyle \text{Therefore: }\:AP \:=\:12-3x\)