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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1210.
We can see that EC is bisector of Angle AEB, lets assume EC intersects AB at point J.By Ceva's theorem AH/HG=(AJ/BJ).(BP/PG)Lets Assume ED extended meets AG at point H', Consider triangle AJG and EDH' as transversal and applying Menelaus' theorem AH'/H'G=(AD/DJ)*(JE/GE) .............(1)If we can prove AH/HG=AH'/H'G, then H and H' coincide, and we get desired result that E,D,H are collinear. Conisder triangle AJE and Triangle EBG Angle AEJ= Angle BEG = x, if We assumne angle EBJ = y then Angle EBG = 180 - y Applying sine law we get JE/GE = AJ/BG We can write AH'/H'G=(AD/DJ)*(AJ/BG)................(2) Applying Menelaus' theorem for Triangle BDP with EC as transversal, we get BG/GP = (BJ/DJ)*(DC/PC)also if we join BC,it is bisectors of Angle DBP hence DC/PC = BD/BP = AD/BP We get BG/PG = (BJ/DJ)*(AD/BP) or BG*DJ =(PG/BP)*AD*BJreplacing BG*DJ in eq 2 , we have AH'/H'G=(AJ/BJ)*(BP/PG) = AH/HG Hence H' and H coincide and E,D,H are collinear.
Correction: if We assumne angle EAJ = y then Angle EBG = 180 - y