How similar triangles can be used to prove that the slope is the same between any two distinct points on a non-vertical line?

Answer 1

Let A, B, C and D be any four points on a non-vertical straight line. Suppose the vertical line from B meets the horizontal line from A at P, and the vertical line from C meets the horizontal line from D at Q. Now, AP and CQ are both horizontal so they are parallel and AC is an intercept. Therefore the corresponding angles BAP and DCQ are equal. BP and DQ are both vertical so they are parallel and BD is an intercept. Therefore the corresponding angles ABP and CDQ are equal. AP and CQ are both horizontal and BP and DQ so that angle APB = angle CQD = 90 degrees. Thus the corresponding angles of triangle ABP are congruent to those of triangle CDQ. Therefore the ratios of their sides is the same. AB/CD = AP/CQ = BP/DQ. In particular, AP/CQ = BP/DQ. By cross multiplication, BP/AP = DQ/CQ or, slope calculated from AB = slope calculated from CD.