## Saturday, October 29, 2016

### Geometry Problem 1280 Quadrilateral, Perpendicular Diagonals, Isosceles Right Triangles, 45 Degrees, Collinear Points

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1280.

1. https://goo.gl/photos/3q9h6djv9nLppAob9

Let M and N are the midpoint of BC and AD
We have MG=MB=MC=ME
And NF=NA=ND=NE
So M and N are the centers of quadrilateral BEGC and AEFD
In qua. BEGC we have ∠CEG=∠CBG= 45 degrees
In qua. AEFD we have ∠FED=∠EAD= 45 degrees
So F and G are located on angle bisector of angle CED => E, F and G are colinnear

2. Problem 1280
Is <AED=AFD=90 so AEFD is cyclic,then <CEF=<FDA=45. But <BEC=<BGC=90 so BEGC is cyclic, then <GEC=<GBC=45=<CEF .Therefore the points E,F and G are collinear.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

3. Suppose that F is not on EG and that F' is the point at which AD subtends 90 on line EG. EF'G are collinear.
Let X be any point on FE extended.

Since BCGE is concyclic < XEB = 45
Hence < XEA = 45 = < F'DA since AEF'D is concyclic.

So F and F' must coincide and E,F,G are hence collinear

Sumith Peiris
Moratuwa
Sri Lanka

4. ∵ The diagonals AC and BD are perpendicular at E
∴ ∠BEC = 90°
∵ AFD and BGC are isosceles right triangle
∴ ∠BGC = 90°
∴ ∠BEC = ∠BGC
∴ B, E, G and C are concyclic
∴ ∠GEC = ∠GBC
∵ AFD and BGC are isosceles right triangle
∴ ∠GBC = ∠GCB
∠GBC + ∠GCB + ∠BGC = 180°
2∠GBC + 90° = 180°
∠GBC = 45°
∴ ∠GEC = 45°
Similarly, ∠DEF = 45°
∵ The diagonals AC and BD are perpendicular at E
∴ ∠CED = 90°
∠DEF +∠CEF = 90°
45° +∠CEF = 90°
∠CEF = 45°
∵ ∠CEG = ∠CEF
∴ The points E, F, and G are collinear