Circle and Chords

The chord CD of a circle center O is perpendicular to the diameter AB.   M_OC is the midpoint of the radius OC.  Line AM_OC is extended to meet the circle at E so that the chord AE goes through the midpoint of OC.  Chord DE intersects chord BC at point P.

Prove that P is the midpoint of BC.

Source: 1995 Russian Math Olympiad questions

Solution

Draw line segments AC and BD.  Now observe that angle ABC is an inscribed angle that is subtended by arc AC, so the measure of angle ABC is half that of arc AC.  Angle ABD = angle ABC by symmetry,  so the measure of angle DBC is twice that of angle ABC, and so angle DBC is congruent to angle AOC.

Triangles DBC and AOC are both isosceles (their bases are chords) with congruent angles, so they are similar.  By the fact that inscribed angles subtending the same arc are congruent, we also have angle BAE congruent to BDE and angle EAC congruent to EDC.

The forgoing shows the similarity of the following pairs of triangles

triangle AOC is similar to triangle DBC
triangle OAM_OC is similar to triangle BDP
triangle CAM_OC is similar to triangle CDP 

From the ratios of corresponding sides of similar triangles,

OM_OC/AM_OC = BP / DP
CM_OC/AM_OC = CP/DP

And since OM_OC=CM_OC, BP/DP = CP/DP, and so BP = CP, proving the statement.

Another approach

There is another way to view this (supplied by Goeff on NRICH).  Take AB a diameter perpendicular to the chord CD.  Now select E an arbitrary point on the circle and let CF be a diameter of the circle.  Now by angle in a semicircle (Thales) FD is parallel to AB.

Now apply Pascal's Theorem on cyclic hexagons to the hexagon AEDFCB with opposite sides color-coded in the diagram.  Opposite sides AB and DF are parallel so meet at infinity.  The line through the points which are the meet, M_OC, of CF and AE, and the meet, P, of CB and ED, must pass through this point at infinity by Pascal's theorem.  In other words, the line through these two points must be parallel to AB.

Now, imagine two additional red lines (not pictured) parallel to AB and DF.  One of them goes through C, and the other through M_OC and P.  The leftmost three red lines are equally spaced, so P is the midpoint of BC.

Internet references

Thales - A triangle in a semicircle is a right triangle.

Related pages in this website

Summary of Geometrical Theorems

The webmaster and author of this Math Help site is Graeme McRae.