Product of the gradients of perpendicular lines proof help

[Theodore Hodson]

Okay I'm having a little trouble understanding a section of this proof about the product of the gradients of perpendicular lines given in my textbook. I'm going to type the proof out but there will be a link at the bottom to an online version of the textbook so you can see the accompanying diagram too.

I understand essentially the whole proof except this one thing it says at the beginning. This bit: 'If AB makes an angle theta with the x-axis then CD makes an angle theta with the y axis.'

So my question is this: Why do the x and y-axis make the same angle (theta) when they intersect AB and CD? What's the reason/proof behind that?

Will be very grateful if anyone can explain this clearly to me:)

Here is the link to my textbook (go to slide 79 - the bit on perpendicular lines at the bottom): http://www.slideshare.net/mobile/Slyscott12/core-maths-for-a-level-3rd-edition-by-lbostock-schandler
Here is the proof typed out:
Consider the perpendicular lines AB and CD whose gradients are m1 and m2 respectively.
If AB makes an angle theta with the x-axis then CD makes an angle theta with the y axis. Therefore triangles PQR and PST are similar.
Now the gradient of AB is ST/PS=m1
And the gradient of CD is - PQ/QR=m2,i.e PQ /QR= - m2

But ST/PS=QR/PQ (since triangles PQR and PST are similar)

Therefore m1= - 1/m2 or m1 *m2 = - 1

 

[robphy]

BPD is a right angle. (Given)

SPQ is a right angle because PS is parallel to the x-axis and QP is parallel to the y-axis.

Now consider angle SPR...

 

[Merlin3189]

1- AB makes an angle θ with y axis. If both lines are rotated 90 deg cw, then AB is now parallel with CD and y-axis is now parallel with x axis, so angle between CD and x-axis is also θ.

2 - Consider the quadrilateral APDO. The corners at O and P are 90 deg by definition. The 4 corners must total 360 , so corners at A and D must total 180. At corner A, the acute angle and obtuse angle with y-axis must also total 180. So the acute angles between the lines and the axes must be equal, and the obtuse angles between the lines and the axes must be equal.

 

[Theodore Hodson]

BPD is a right angle. (Given)

SPQ is a right angle because PS is parallel to the x-axis and QP is parallel to the y-axis.

Now consider angle SPR...

Okay I think I might have it. So going from what you said:

∠ BPD=90° and ∠SPQ=90°

From the diagram ∠BPS+∠SPR =∠BPD
then ∠BPS +∠SPR = 90°

So ∠BPS = 90°-∠SPR

Also, as ∠SPR +∠RPQ =∠SPQ then

∠SPR +∠RPQ = 90°

So ∠RPQ = 90° - ∠SPR

Therefore ∠RPQ = ∠BPS

And in the book they've just marked ∠RPQ and ∠BPS down as theta. Is this right?