ABC is a triangle. A circle through B and C meets the side AB again at C' and meets the side AC again at B'. Let H be the orthocenter of ABC and H' the orthocenter of AB'C'. Show that the lines BB', CC' and HH' are concurrent.
Solution
Let P be the intersection of BB' and CC'. We have to show that angle BPH = angle B'PH'.
Since BC'B'C is cyclic, we have ∠BB'C = ∠BC'C. Hence ∠PBH = 90o - ∠BB'C = 90o - ∠BC'C = ∠PCH. Take E so that PCEH is a parallelogram. Then ∠PCH = ∠CHE. Take D so that BPCD is a parallelgram. Then BD is equal in length and parallel to PC, which is equal in length and parallel to HE. So BDEH is also a parallelogram, so BH = DE. The other parallelograms give BP = DC and PH = CE, so BPH and DCE are congruent. Hence ∠PBH = ∠CDE. So ∠CDE = ∠CHE and CHDE is cyclic. So ∠BPH = ∠DCE (congruent triangles) = ∠DHE (CHDE cyclic). But BHED is a parallelogram, so ∠DHE = ∠HDB. Hence ∠PBH = ∠HDB.
Now ∠AB'C' = 180o - ∠C'B'C = ∠C'BC (cyclic) = ∠ABC (same angle). So ABC and AB'C' are similar. Hence HBC and H'B'C' are similar. PCB and PB'C' are similar. But PCB and DBC are similar, so DBC and PB'C' are similar. Hence BHCD and B'H'C'P are similar. Hence BHD and B'H'P are similar. So ∠HDB = ∠H'PB' and hence ∠PBH = ∠H'PB', as required.
© John Scholes
jscholes@kalva.demon.co.uk
9 Oct 2002