Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 69Square Inscribed in a Triangle. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
Let the vertices of square are PQRS and BM perpendicular is drawn to BC. area triangle APM=(1/2).s.AMarea of triangle SMC=(1/2).s.MCarea of triangle PMS=(1/2).s.sarea of triangle BPS=(1/2).s.(h-s)area of whole triangle=(1/2).b.hnow1/2).s.AM+(1/2).s.MC+(1/2).s.s+(1/2).s.(h-s)=(1/2).b.hand AM + MC=bwhich on simplifying prove the relation
With P, Q, R and S vertices of the square and k be the height of the triangle ΔSBR, we have ΔABC ~ ΔSBR. Therefore,From h/k = b/s, we have:s = bk/h = b(h-s)/h = (bh-bs)/hand sh + bs = bh or s = bh/(b+h)QED
Let PQRS be the square.Area of triangle ABC= Area of triangle BSR + Area of Trapezium ACRS.(1/2)bh = (1/2)s*(h-s) + (1/2)(b+s)* (h-s)bh = (h-s)(2s+b)on simplification we get the result
Area of Tr. ABC is comprised of the areas of 3 Tr.s and the square. So s^2 + 1/2s(h-s) + 1/2s(b-s) = 1/2 bhFrom which the result followsSumith PeirisMoratuwaSri Lanka
Or from similar triangles (h-s)/h .= s/b and the result follows