## Friday, October 18, 2013

### Geometry Problem 932: Square, Midpoint, Diagonal, Ratio 3:1, Angle Measure, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 932. 1. This problem is similar to problem 879. refer to links below for solution
http://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

2. 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°

3. A graphical solution: http://bleaug.free.fr/gogeometry/932.png

bleaug

4. Let A(0,0), B (0,4a), C(4a,4a), D(4a,0) be the vertices.
Clearly 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.

5. Join 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

6. BF = DF (Tr.s AFD, AFB being congruent)

Perpendicular 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