ABCD is a cyclic quadrilateral. A' is the orthocenter (point where the altitudes meet) of BCD, B' is the orthocenter of ACD, C' is the orthocenter of ABD, and D' is the orthocenter of ABC. Prove that ABCD and A'B'C'D' are congruent.
Solution
Let A" be the centroid of BCD, B" the centroid of ACD, C" the centroid of ABD and D" the centroid of ABC. We claim that A"B"C"D" is similar to ABCD and 1/3 the size.
Take any two points of A, B, C, D, wlog A and B. Let M the midpoint of the other two. Then A" is 1/3 the way along MB and B" is 1/3 the way along MA. Hence A"B" = AB/3. Similarly, all the other distances in A"B"C"D" are 1/3 the corresponding distances in ABCD, which establishes the claim.
Now consider the Euler line of the triangle BCD. Its circumcenter is O, its centroid is A'' and its altitudes meet at A'. So these three points lie on the Euler line and OA' = 3OA". Similarly, OB' = 3OB", OC' = 3OC" and OD' = 3OD". So A'B'C'D' is similar to A"B"C"D" and 3 times the size. Hence it is congruent to ABCD.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002
Last corrected/updated 26 Jan 04