tag:blogger.com,1999:blog-6933544261975483399.post538023192716069184..comments2024-10-13T09:45:37.126-07:00Comments on GoGeometry.com (Problem Solutions): Geometry Problem 1395: Triangle with three rectangles on the sides, Midpoints, Perpendicular lines, ConcurrencyAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-18405298046671734572018-10-28T15:36:20.325-07:002018-10-28T15:36:20.325-07:00https://photos.app.goo.gl/3a76E7ztEMNkD1fw6
Let u...https://photos.app.goo.gl/3a76E7ztEMNkD1fw6<br /><br />Let u, v, x, y , w, z are angles of triangles HCF, BDG and AEJ as shown in the sketch<br />In triangle HCF draw altitude Q’Ct<br />Note that ∠ (BCt)= u and ∠ (ACt)= v ( angles formed by perpendicular lines)<br />We also have ∠ (MKQ)= ∠ (BCt)= u ( angles formed by parallel lines)<br />And ∠ (LKQ)= ∠ (ACt)= v<br />Similarly we also have x, y,w, z inside triangle KLM as shown <br />We have sin(u)/sin(v)= CH/CF , sin(y)/sin(x)= BD/BG and sin(w)/sin(z)= AJ/AE<br />Multiply 3 expressions side by side and note that ABDE, BCFG and ACHJ are the rectangles<br />We have sin(u)/sin(v) * sin ( y)/sin(x) * sin(w)/sin(z)= 1<br />So KQ , NL and MP are concurrent per Ceva’s theorem<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com