Online Geometry theorems, problems, solutions, and related topics.
Proposed ProblemClick the figure below to see the complete problem 317.See more:Complete Problem 317Level: High School, SAT Prep, College geometry
Denoting AH & PC by c & d resply. and noting the fact that all rt. angled triangles within Tr. ABC are similar to it, we can say that c/(c+a)= a/x while d/(d+b)= b/x. Further, c/a = AB/BC and d/b = BC/AB or c/(c+a)= AB/(AB+BC) and d/(d+b)= BC/(AB+BC). This gives us, a/x + b/x = c/(c+a) + d/(d+b)= AB/(AB+BC) + BC/(AB+BC) = 1. In other words, x = a + b QED.
tr JKE and tr EBF are similar ( an JEK = an BFE )=> EB/BF = a/x-a tr EBF and MFN are similar ( an BFE = an FNM )=> EB/BF = x-b/bfrom above => a/x-a = x-b/b => x = a + b