Question 1182584
<font color=black size=3>
Refer to the diagram below
<img width="50%" src = "https://i.imgur.com/wolstjT.png">


Triangle ABC is a right triangle. The 90 degree is at angle A.


Sides AB and BC are p and q units long respectively. Where p & q are positive real numbers. We'll make p < q.


Notice how the slope of line BC is -q/p
rise = -q
run = p
The slope is negative since we're going downhill as we move from left to right (from C to B).


So we can consider m1 = -q/p.
We'll use this later.


----------------------------


Triangle CDE is also a right triangle with sides CD = p and DE = q


Notice how triangle CDE is a rotated version of triangle ABC. Triangle ABC and DCE are congruent triangles.


For either right triangle (ABC or DCE), the acute angles add to 90. Since we're dealing with identical triangles, this means that the acute angles DCE and ACB are also complementary.


Put another way:
angle ACB = angle DEC
(angle DCE) + (angle DEC) = 90
(angle DCE) + (angle ACB) = 90


Now since ACD is a straight angle, we can then say,
(angle DCE) + (angle ECB) + (angle ACB) = 180
[ (angle DCE) + (angle ACB) ] + (angle ECB) = 180
[ 90 ] + (angle ECB) = 180
angle ECB = 180 - 90
angle ECB = 90


This proves that angle ECB is a right angle, and it further shows segment BC is perpendicular to EC.


Note that the slope of line EC is rise/run = p/q
We'll call this m2, so,
m2 = p/q


----------------------------


Let's multiply the slopes m1 and m2


m1*m2 = (-q/p)*(p/q)
m1*m2 = (-q*p)/(p*q)
m1*m2 = -1
The p's and q's cancel when we divide them. 


This concludes the proof. 
</font>