Distance from a point to a line the perpendicular "R". Draw the vertical line through "P" and label its intersection with the given line "S". At any point "T" on the line, draw a right triangle "TVU" whose sides are horizontal and vertical line segments with hypotenuse "TU" on the given line and horizontal side of length |"B"| (see diagram). The vertical side of ∆"TVU" will have length |"A"| since the line has slope -"A"/"B". ∆"PRS" and ∆"TVU" are similar triangles, since they are both right triangles and ∠"PSR" ≅ ∠"TUV" since they are corresponding angles of a transversal to the parallel lines "PS" and "UV" (both
Triangle inequality angles satisfy: Therefore, and so, in particular, That means side opposite angle is shorter than side opposite the larger angle . But . Hence: A similar construction shows , establishing the theorem. An alternative proof (also based upon the triangle postulate) proceeds by considering three positions for point : (i) as depicted (which is to be proven), or (ii) coincident with (which would mean the isosceles triangle had two right angles as base angles plus the vertex angle , which would violate the triangle postulate), or lastly, (iii) interior to the right triangle between points and (in which case angle