Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view more details of problem 1280.
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.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view more details of problem 1280.
https://goo.gl/photos/3q9h6djv9nLppAob9
ReplyDeleteLet 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
Problem 1280
ReplyDeleteIs <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
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.
ReplyDeleteLet 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
∵ The diagonals AC and BD are perpendicular at E
ReplyDelete∴ ∠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