Midpoint of a Straight line

Which of the below two is easier to convince yourself of?

a) the angle-side-angle theorem of triangle congruence

b) relative proportions of a line remain consistent with its projection (shadow).

i know it ought to be b).... but on a muddled up day...  you invent some muddle up logic ...

Don't ask me why but i had to calculate the midpoint of a straight line between two points : A (x1, y1) and B: (x2, y2) on the cartesian plane :-)

There is the well known formula for the midpoint, if we call it C : ( (x1+x2)/2, (y1+y2)/2 )

Though it is visually convincing to see that the x co-ordinate and y co-ordinate of the mid point C fall midway between the x and y co-ordinates of the point A and B, i was trying to build some logical reasoning behind it, if not the proof. Looked around a bit online, but couldn't really find a satisfactory explanation and hence wrote one for myself.

Let's take a look at the below example with points A (1, 4) and B (7, 2) whose midpoint lies at C (4, 3)

Its easy to see this by taking the projections of points A, B, C on X and Y axes and realizing that the X-cord of C is exactly half way between the X-coords of A and B. And likewise with the Y-Coords...

But it seems difficult to put this into a logically tight expression. 

How else can we deduce the midpoint in a sound way? Lets define the midpoint again.. It is a point that lies exactly half way between the end points of a finite line segment.

So, if we consider the AB as the line segment, and C as the midpoint, length of AC is same as length of CB. Lets denote this length AC = CB = m. 

Let's look at the above same picture but with a different perspective now, repurposed as below:

The triangles AOC, and CQB are congruent because:

a)  ∠AOC = ∠CQB = 90

b) ∠OCA = ∠QBC  Why? Because, A straight line (AB) makes the same angle with two parallel lines (OC and QB).

c) Length of sides AC = CB = m 

d) Additionally, ∠CAO = ∠BCQ (from a and b)

So, using angle-angle-side or angle-side-angle theorems, we can conclude that triangles AOC and CQB are congurent.

Hence:

1) side AO = CQ Which means Y-coord of C is exactly halfway between the Y-cords of points A and B, given that Q's Y-Coord is same as B's Y-Coord.

2) side OC = QB. Which means X-coord of C is exactly halfway between the X-coords of points A and B, given that O and A have same X-coords, and C and Q have same X-coords.

Ofcourse, here we took help of angle-side-angle theorem for our deduction.