Prove every segment has a midpoint.

Unfortunately I do not have the definition yet of isosceles triangles. All I have is SSS and SAS. I also do not have right angles. But I do have perpendicular lines and bisectors.

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    $\begingroup$ You may have a look at page (3/8) from this link $\endgroup$ – Arashium Feb 18 '15 at 19:26

If we are in euclidean geometry you forget the circle,infact if we have a segment AB,
let's take the circle C1 with center in A and radius the lenght of AB and the circle C2
with center in B and radius AB,if we intersect these 2 circle we obtain two different points,
let's call them P and Q,they have the same distance from the A and B,if we call r the straight line
for P and Q,r is the AXIS of the segment,in short words the locus of the points with the same distance from A and B,and the intersection between r and the segment AB is the middle point.
The straight r is perpendicular to AB then this intersection always exists.
Here it is an animation that explain the construction.


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