tag:blogger.com,1999:blog-6933544261975483399.post7092007288490337102..comments2024-03-19T00:02:30.728-07:00Comments on Go Geometry (Problem Solutions): Dynamic Geometry 1460: Newton-Gauss Line, Complete Quadrilateral, Midpoints of DiagonalsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-44903347897940578832020-05-07T11:47:32.852-07:002020-05-07T11:47:32.852-07:00https://photos.app.goo.gl/fuXMGA8D5W6nwNzV8
Defin...https://photos.app.goo.gl/fuXMGA8D5W6nwNzV8<br /><br />Define points K, P, Q as midpoint of CF, DF, CD ( see sketch)<br />Note that T, K, P are collinear <br />Same as W,Q,P and N, Q, K<br />Consider triangle CDF and secant ABF, we have<br />EC/ED x BC/BF x AD/AF = 1…..(1)<br />But EC/ED= TK/TP ; BC/BF= WP/WQ ; AD/AF=NQ/NK<br />Replace it in ( 1) then we have TK/TP x WP/WQ x NQ/NK= 1<br />So N, W, T are colinear per Menelaus’s theorem <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com