## Wednesday, May 4, 2016

### Geometry Problem1210. Circle, Tangent Line, Secant, Chord, Collinear Points

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.

#### 2 comments:

1. 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*BJ
replacing 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.

2. Correction: if We assumne angle EAJ = y then Angle EBG = 180 - y