Problem:
(*) Let O and H be the circumcenter and the orthocenter of an acute triangle ABC. Points M and D lie on side BC such that BM=CM and ∠BAD=∠CAD. Ray MO intersects the circumcircle of triangle BHC in point N. Prove that ∠ADO=∠HAN.
Solution:
Set ∠CAB=A,∠ABC=B, and ∠BCA=C. Because H is the orthocenter, we have ∠HBC= 90∘−C and ∠HCB=90∘−B. In triangle BHC, we have ∠BHC=180∘−∠HBC−∠HCB= B+C. Because BHNC is cyclic, we have ∠BNC=∠BHC=B+C. Extend segment AD through D to meet the circumcircle (denoted by ω ) of triangle ABC at P. It is clear that P is the midpoint of minor arc \overparen{B C} (of ω ) and O,M,P all lie on the perpendicular bisector of segment\
BC. In particular, BPCN is a kite with symmetry axis PN. Because ABPC is cyclic, we have ∠BPC=180∘−∠BAC=B+C=∠BNC. We can further conclude that BPCN is a rhombus, implying that line BC is the perpendicular bisector of segment NP, and so DN=NP and ∠DPN=∠DNP.
Set x=∠HAP. Because AH∥OP, we have ∠DNP=∠DPN=∠HAP=x. Because O is the circumcenter of triangle ABC, we have ∠AOC=2B and ∠CAO=∠ACO=90∘−B. Because H is the orthocenter of triangle ABC, we have ∠BAH=90∘−B. Because ∠BAH=90∘−B=∠CAO, ∠BAC and ∠HAO share common angle bisector AD; that is,
∠DNP=∠DPN=∠HAP=∠OAP=∠OAD=x
Consequently, we have
∠ADO=∠ADN−∠ODN=∠DNP+∠DPN−∠ODN=2x−∠ODN
and
∠HAN=∠HAO−∠OAN=∠HAP+∠OAP−∠OAN=2x−∠OAN
It suffices to show that ∠ODN=∠OAN, which is clearly true because ADNO is cyclic as ∠DNP=∠OAD=x.
Alternate solution (by Titu Andreescu and Cosmin Pohoata). The key idea is to prove that ADNO is cyclic. Once this is proven, the problem follows by noticing that ∠ADO=∠ANO= ∠HAN, where the latter holds due to the fact that ON∥AH.
To prove the concyclicity, one can simply use Power of a Point. First, one has to construct P as in the first solution, and notice that M is the midpoint of segment PN. This follows from the fact that the reflection of H across line BC lies on the circumcircle Ω of △ABC. This implies that the circumcircle of △BHC is the reflection of Ω across line BC, so line BC must indeed bisect PN by symmetry. Next, let O′ denote the orthogonal projection of O on AD. Clearly OO′DM is cyclic, so Power of a Point yields PM⋅PO=PD⋅PO′. But O′ is the midpoint of PA, so PO′=PA/2. Since PM=PN/2, this yields
PNâ‹…PO=PDâ‹…PA
which by Power of a Point gives the concyclity of ADNO. This completes the proof.
The problems on this page are the property of the MAA's American Mathematics Competitions