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

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

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## Saturday, April 4, 2015

### Problem 1107: Triangle, Circumcircle, Circle, Radius, Tangent, Chord, Congruence

Labels:
chord,
circle,
circumcircle,
congruence,
radius,
tangent,
triangle

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ReplyDeleteLet circle D cut AB at E and F

Since DC^2=DA.DB => Circle D is Apollonius circle of AB

And PE and CE are bisectors of angles APB and ACB

So PA/PB=EA/EB=CA/CB (1)

Triangles PBC similar to PC1B1 => B1C1= (PC1/PB). BC…..(2)

Triangles PAC similar to PC1A1=> C1A1= (PC1/PA). CA …..(3)

Divide (2) to (3) and using (1) we get B1C1= C1A1

Let AB cut circle D at Q.

ReplyDeleteLet < BCD = p, <BPD = q, < QPC = r and < QCP = s

Then <CAD = p and < PAD = q since DP (=DC) is a tangent to circle APB at P.

Further < PDA = 2s and < CDA = 2r so that

< PQD = < QPD = 90-s …..(1) and

< QCD = CQD = 90-r…….(2)

From (2) < ACQ = 90-p-r …..(3)

From (1) < APQ = 90-q-s ……(4)

Now < ABB1 = 2s+q & …..(5)

And < CBA = 2r+p ……(6)

Further < PCB = 90-p-r-s from which

< A1B1C1 = 90-p-q-r-s …..(7) since < PCB = < PB1C1 and < BB1A = BAA1 = q

Also < B1A1C1 = < PA1C1 - <PA1B1 = < ACC1 - < ABB1 = (90-p-r+s) – (2s+q) =

90-p-q-r-s…(8)

From (7) and (8), < A1B1C1 = < B1A1C1 and therefore C1A1 = C1B1

Sumith Peiris

Moratuwa

Sri Lanka