perpendicular lines proof.?

Question

Im a bit stuck on this proof example in the book, I will type it out and then show where I am stuck.

It says

Consider perpendicular lines AB and CD whose gradients are m1 and m2 respectively

If AB makes an angle (theta) with the x axis and CD makes an angle (theta) with the right axis therefore triangle PQR and PST are similar.

(To ellaborate seen as ya cant draw on this there are now 2 triangles on each perpendicular line PST is on line ab and has the angle theta with the x axis where it bisects line CD, PS is the height TP the hypotenuse and TS the base.

The other triangle is on line CD theta forms from the point where it bisects BA down parrallel to the y axis to Q and then to R.

So Gradient of AB is ST/PS =m1 (I get that)

and the gradient of CD is -PQ/QR i.e PQ/QR = -m2 (I get that)

**But ST/PS = QR/PQ (I get that because in similar triangles the sides have to divide in the same ratio)

Therefore m1 = - 1/m2 (Im lost here) How did they get to this from **

Answer 1

***Edited because I made a typo
Well, if you look closely, you already have the answer before your eyes. I think you accidentally overlooked this.

You said that ST/PS = m1
You said that PQ/QR = -m2
You said that ST/PS = QR/PQ
Well, ST/PS = m1
PQ/QR = -m2 therefore QR/PQ = -1/m2
Now just substitute that into ST/PS = QR/PQ and you get
m1 = -1/m2
You got it dude :)
-The Mad Scientist

Answer 2

Substitute pieces from the statements above the one you don't get.

m1 = ST/PS
**But ST/PS = QR/PQ (I get that because...

so replace m1 for ST/PS -->m1 = QR / PQ
(remember this step)

m2 = -PQ / QR
1/m2 = -QR / PQ
-1/m2 = QR / PQ

remember m1 = QR / PQ from above? Replace that too.
-1/m2 = m1
m1 = -1/m2 (flip it around to be the same as the answer)


Read another way, this proof says that the slopes of perpendicular lines are negative reciprocals of each other.
e.g. a line of slope 2 is perpendicular to a line of slope -1/2