如图,H为三角形ABC垂心、O为其外心,DE过H且AD=AE,M为BC中点,O'为三角形ADE外心,证OO'平行于HM

Question

如图,H为三角形ABC垂心、O为其外心,DE过H且AD=AE,M为BC中点,O'为三角形ADE外心,证OO'平行于HM

Answer 1

延长AO交圆O于A',则A'C垂直于AC.
BH=2R(COS∠ABC)=2R(COS∠AA'C)=A'C
由M是BC中点,故HA'过M.设BH交AC于B'
由于AD=AE,∠B'HE=90°-∠AED=∠BAC/2,∠CHE=∠AED-∠ACH=∠BAC/2.
所以HE平分∠B'HC.
所以B'E/EC=HB'/HC=COS∠BAC=AH/AA'
设AO'交HA'于Q,AO'平分∠BAC,
而∠BAH=∠CAA'=90°-∠ABC,所以∠AO'平分∠HAA'
故HQ/QA'=AH/AA'=B'E/EC,
但是BB'//A'C所以QE//BB',QE垂直AE,所以AQ是圆O'直径,
AO'=O'Q=OA'
故OO'//A'Q,OO'//HM