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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.
Denote inscribed angles of arcs PTa, PTc, and TcTb to be a, b, and c respectively.a1=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
https://goo.gl/photos/8vk1AQ8RWkqjdc3Q9Let PQ is the diameter of circle O and R= radius of circle ODefine a as ∠ (PTaB)= ∠ (PQTa). b=∠ (BTcP)= ∠ (PQTc). c= ∠ (ATbTc)= ∠ (TbQTc)See sketch for detailsWe haveSin(a)=PTa/2.R Sin(b)=PTc/2RSin(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.TbTcPlug these values in above equation we will get the result