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

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

Click the figure below to see the complete problem 1024.

## Monday, June 16, 2014

### Geometry Problem 1024: Contact, Gergonne Triangle, Point on an arc of Incircle, Perpendicular, Distances

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Denote inscribed angles of arcs PTa, PTc, and TcTb to be a, b, and c respectively.

ReplyDeletea1=PTa*sin(a)=(PTa)^2/(2R)

b1=PTb*sin(b+c)=(PTb)^2/(2R)

c1=PTc*sin(b)=(PTc)^2/(2R)

Plugging these into equation to be proved yields Ptolemy's Theorem in quadrilateral PTaTbTc.

Can someone post a solution to this question, I'm having a difficult time proving it

ReplyDeletehttps://goo.gl/photos/8vk1AQ8RWkqjdc3Q9

ReplyDeleteLet PQ is the diameter of circle O and R= radius of circle O

Define a as ∠ (PTaB)= ∠ (PQTa)

. b=∠ (BTcP)= ∠ (PQTc)

. c= ∠ (ATbTc)= ∠ (TbQTc)

See sketch for details

We have

Sin(a)=PTa/2.R

Sin(b)=PTc/2R

Sin(b+c)= PTb/2R

a1=PTa*sin(a)=(PTa)^2/(2R) => PTa= sqrt(2.R.a1)

b1=PTb*sin(b+c)=(PTb)^2/(2R)=> PTb= sqrt(2.r.b1)

c1=PTc*sin(b)=(PTc)^2/(2R) => PTc= sqrt(2.R.c1)

Apply Ptolemy’s theorem in quad. PTaTbTc

we have TaTc. PTb= PTc.TaTb+ PTa.TbTc

Plug these values in above equation we will get the result