Proposed Problem

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Complete Problem 317

Level: High School, SAT Prep, College geometry

## Thursday, July 9, 2009

### Problem 317: Right triangle and Inscribed Squares

Labels:
inscribed,
right triangle,
square

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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.

ReplyDeletetr JKE and tr EBF are similar ( an JEK = an BFE )

ReplyDelete=> EB/BF = a/x-a

tr EBF and MFN are similar ( an BFE = an FNM )

=> EB/BF = x-b/b

from above => a/x-a = x-b/b

=> x = a + b