Geometry Problem. Post your solution in the comment box below.

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

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## Saturday, October 29, 2016

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

Labels:
45 degrees,
collinear,
diagonal,
isosceles,
perpendicular,
quadrilateral,
right triangle

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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