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

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

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## Saturday, February 27, 2016

### Geometry Problem 1191: Circle, Chords, Diameter, Congruence, Midpoint, Collinearity, Bisector

Labels:
angle bisector,
chord,
circle,
congruence,
diameter,
midpoint,
secant

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1. Consider hexagon EGDCHF

ReplyDeletewe have P= EF∩DC, O=HC∩EG,J=DG∩HF

so P, O, J are collinear per Pascal’s theorem

2. In circle O , since PA=PB => JP⊥ AB

3. Observe that FJPK and JPDL are cyclic quadrilateral

So ∡ (KJP)= ∡ (KFP)= ∡ (EDC)= ∡ (PJL)

So JO bisect ∡ (KJL)

4. KJL is isosceles tri. => JK=JL and KA=BL

5. Since KO=LO => power of K or L to circle O have the same value= KC.KF= LE.LD

Good work Peter in noting to apply Pascal to the cyclic hexagon and showing that J,O,P are collinear after which the rest is fairly straightforward.

DeleteOn the last point, KF.KC = KA.KB = BL.LA (since KA = BL) = LD.LE