## Saturday, January 31, 2009

### Elearn Geometry Problem 238: Square, Midpoints, Lines, Congruence

See complete Problem 238 at:
gogeometry.com/problem/p238_square_midpoints_congruence.htm

Level: High School, SAT Prep, College geometry

1. It is easy to prove that the measure of AGF is 90(traingles ABE and BCF are congruents). Let H be the intersection of BF and AD. FD= AB/2, FD is parallel to AB then, AD=DH, so GD is median of ortogonal triangle AGH. So GD = AH /2 = AD = AB.
There is another solution with calculus.

2. Here's another by analytical geom.: Let A be(0,0),B:(0,a),C:(a,a)& D:(a,0. So E:(a/2,a) & F:(a,a/2)and BF:x/2a + y/a = 1, AE:y = 2x. This makes G:(2a/5,4a/5) and GD^2=b^2=(2a/5-a)^2+(4a/5)^2 = a^2. Hence a=b.
I want to know how this can be solved by calculus. Ajit: ajitathle@gmail.com

3. We observe that AGFD is a cyclic quadrilateral.
also, angle BFC = AFD and AFD = AGD.

4. http://www.mathematica.gr/forum/viewtopic.php?f=20&t=32650&p=151226

5. ABE,BCF,ADF are congruent right triangles.

So < BAE = < FBC = < FAD = p say.

AE & BF are hence perpendicualar and AGFD is concyclic and hence < GAF = 90-2p & so < GFA = 2p = < ADG

Therefore < GAD = < AGD = 90-p and so b = GD = AD = a

Sumith Peiris
Moratuwa
Sri Lanka

6. - b is the length of chord GD intercepted by angle <DAG in circle ADFG of diameter AF
- a is the length of chord AB intercepted by angle <AEB in circle ABE of diameter AE
-but AE = AF and <AEB = <DAG hence b = a

7. See the drawing

- Triangle ABE is congruent to triangle BCF by rotation of pi/2 => < AGF=90°
- Define H as the middle of AB and K as the middle of AG
- By construction HD//BF
- AH=HB=>AK=KG
- < AGF=90° => < AKD=90°
- triangle ADG has its height in the middle of AG
- => ADG is isosceles in D => DA=DG
- Therefore a=b

8. Draw DLM, L on AG, M midpoint of AB. Tr AML~ Tr GLD with coefficient 0.5

9. See the drawing

- ΔABE is congruent to ΔBCF by rotation of pi/2 => ∠ AGF=90°
- Define H as the middle of AB and K as the middle of AG
- By construction HD//BF
- AH=HB=>AK=KG
- ∠ AGF=90° => ∠ AKD=90°
- ΔDKG is congruent to ΔDKA (SAS)
- =>DG=DA
- Therefore a=b

10. Triangle BGE ~ Triangle BCF
BG:GE=2:1
GE=c, BG=2c
----------------------------------------
By Pyth. Theorem (Triangle BEG)
BE^2=BG^2+GE^2=5c^2
BE^2=(a/2)^2
=>a^2=20c^2
----------------------------------------
By Pyth. Theorem (Triangle BAE)
BE^2+AB^2=AE^2
AE^2=1.25a^2=25c^2
AE=5c
AG=AE-GE=4a
-----------------------------------------
cos0

11. Solution 2

Triangles ABE, AFD & BCF are congruent SAS
So < BFC = < AEB and hence EGCF is concyclic
BF is thus perpendicular to AE and so ADFG is also concyclic
Therefore < BFC = < GAD = < AFD = < AGD
So AD = GD and a = b

Sumith Peiris
Moratuwa
Sri Lanka