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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 1023.
This problem can be solved by similiarty of quadrilaterals:Step 1:C1PB2Tc ~ B2PA1Ta (3 equal angles -> 2 pairs of 90 and we have BTc = BTa (Because of the tangents) ---> b2^2 = a1*c1Notice that we have to prove that a1*b1*c1 = a2*b2*c2 /we will devide by b2^2 = a1*c1Thus we will have to prove that b1= a2*c2/b2 ---> b1*b2= a2*c2Step 2:Let <CTaTb = <CTbTa = a = < TaTcTb (Because of the tangents rules and equality)Thus < C2TbB1 = < B2TcA2 = 180 - a (as mentoned above)Notice again that C2TbB1P ~ PB2TcA2 ( 3 equal pairs of angles)From simple proportion we get : b1/a2 = c2/b2 Q.E.D as we had to prove in Step 1.
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a1/b2=sin(PTbTa)/sin(PTaTc) due to PA1TaB2 cyclic and tangent anglesb1/c2=sin(PTaTb)/sin(PTbTa) due to PC2TbB1 cyclic and tangent anglesc1/a2=sin(PTaTc)/sin(PTaTb) due to PA2TcC1 cyclic and tangent anglesMultiply top three equations to get equation to be proved (a1b1c1)/(a2b2c2)=1
Use the lemma from https://www.facebook.com/photo.php?fbid=777498538997721&set=p.777498538997721&type=1&theaterBest regards,S. Fulger
Problem 1023According to the problem 1020 we have PC_2^2=PA_1.PB_1, PB_2^2=PA_1.PC_1, PA_2^2=PC_1.PB_1.By multiplying by members we have PA_2^2. PB_2^2. PC_2^2=PA_1^2.PB_1^2.PC_1^2, or PA_1. PB_1. PC_1=PA_2.PB_2.PC_2.APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE