A formal proof of the theorem is as follows: Let the
perpendiculars and be dropped from the point on the straight lines and respectively. Similarly, let and be dropped from the point perpendicular to the straight lines and respectively. Since :: \triangle MXX' \sim \triangle MYY', : {MX \over MY} = {XX' \over YY'}, :: \triangle MXX
\sim \triangle MYY, : {MX \over MY} = {XX
\over YY}, :: \triangle AXX' \sim \triangle CYY'', : {XX' \over YY''} = {AX \over CY}, :: \triangle DXX'' \sim \triangle BYY', : {XX'' \over YY'} = {DX \over BY}. From the preceding equations and the
intersecting chords theorem, it can be seen that : \left({MX \over MY}\right)^2 = {XX' \over YY' } {XX
\over YY}, : {} = {AX \cdot DX \over CY \cdot BY}, : {} = {PX \cdot QX \over PY \cdot QY}, : {} = {(PM-XM) \cdot (MQ+XM) \over (PM+MY) \cdot (QM-MY)}, : {} = { (PM)^2 - (MX)^2 \over (PM)^2 - (MY)^2}, since . So, : { (MX)^2 \over (MY)^2} = {(PM)^2 - (MX)^2 \over (PM)^2 - (MY)^2}. Cross-multiplying in the latter equation, : {(MX)^2 \cdot (PM)^2 - (MX)^2 \cdot (MY)^2} = {(MY)^2 \cdot (PM)^2 - (MX)^2 \cdot (MY)^2} . Cancelling the common term : { -(MX)^2 \cdot (MY)^2} from both sides of the equation yields : {(MX)^2 \cdot (PM)^2} = {(MY)^2 \cdot (PM)^2}, hence , since MX, MY, and PM are all positive, real numbers. Thus, is the midpoint of . Other proofs exist, including one using
projective geometry. ==History==