Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Wednesday, May 4, 2016

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

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We can see that EC is bisector of Angle AEB, lets assume EC intersects AB at point J.

ReplyDeleteBy 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.

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

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