Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 932.
Friday, October 18, 2013
Geometry Problem 932: Square, Midpoint, Diagonal, Ratio 3:1, Angle Measure, Congruence
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This problem is similar to problem 879. refer to links below for solution
ReplyDeletehttp://gogeometry.blogspot.com/2013/05/problem-879-square-midpoint-diagonal.html
http://img208.imageshack.us/img208/6976/problem897.png
angle(FED)=angle(EDF)= 45 so angle(EFD)=90
As shown by Peter, it's easy to show that FE=FD and ///ly FB=FE. Thus, a circle drawn with F as the centre and radius=FE passes through points D & B as well. Therefore, /_DFE=2*/_DBE =90°
ReplyDeleteA graphical solution: http://bleaug.free.fr/gogeometry/932.png
ReplyDeletebleaug
Let A(0,0), B (0,4a), C(4a,4a), D(4a,0) be the vertices.
ReplyDeleteClearly E(2a,4a) and F(a,a) are the other points.
EF^2 = 10a^2 = DF^2, ED^2 = 20a^2.
ED^2 = EF^2+ DF^2.
By Pythagoras EFD is a right angle.
Let m(ADF)=x=> m(AFD)=135-x
ReplyDeleteJoin BF and the triangles AFD and AFB are congruent (SAS) => BF=FD and m(BFA)=135-x -----------(1)
Join BD,DB and form the triangle BED
Observe that in the triangle BED, BE/BD = AF/AD and m(DAF)=m(DBE)=45
=> BED is similar to AFD and m(BFD) = 135-x => 1/2m(BFD) ---------(2)
From (1) and (2), F is the circumcenter of triangle BED and m(EFD)=2.m(EBD)=90
BF = DF (Tr.s AFD, AFB being congruent)
ReplyDeletePerpendicular bisector of BE passes thro F, so FE = FB
Hence < CDF = < FBE = < FEB so CEFD is concyclic and the result follows
Sumith Peiris
Moratuwa
Sri Lanka