All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Rhombus problem, with tangents and parallel (Posted on 2013-01-13)
ABCD is a rhombus. Take points E, F, G, H on sides AB, BC, CD, DA respectively so that EF and GH are tangent to the incircle of ABCD. Show that EH and FG are parallel.

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution Comment 1 of 1
` It is easy to show that EH and FG are parallel if and only if triangles HAE and FCG are similar. Let O be the intersection of diagonals AC and BD(also, it's the center of the incircle). Let E',F', G', and H' be the points of tangency of theincircle with the sides AB, BC, CD, and DArespectively. Let I and J be the points of tangency of the incircle with EF and GH respectively. In the following a string of threecapital letter denotes an angle unless specifiedotherwise. Let the twelve angles about O be     u = H'OH = HOJ    v = JOG  = GOG'    z = G'OC = COF'    x = F'OF = FOI    y = IOE  = EOE'    z = E'OA = AOH'. Clearly, the following holds     u + v + z = x + y + z = 90.              (1) With algebra and a trig. identity (1) implies     tan(x)+tan(z)     tan(v)+tan(z)   --------------- = ---------------.        (2)    tan(z)+tan(u)     tan(z)+tan(y)   Multiplying both sides of (2) with r/r (wherer is radius of the incircle) gives     r*tan(x)+r*tan(z)     r*tan(v)+r*tan(z)   ------------------- = -------------------       r*tan(z)+r*tan(u)     r*tan(z)+r*tan(y)                        or     |EE'| + |E'A|     |GG'| + |G'C|   --------------- = ---------------       |AH'| + |H'H|     |CF'| + |F'F|                  or     |EA|     |GC|   ------ = ------.       |AH|     |CF|   This and HAE = FCG implies the trianglesare similar. QED`

 Posted by Bractals on 2013-01-14 13:07:32

 Search: Search body:
Forums (0)