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

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click the figure below to view more details of problem 1234.

## Saturday, July 9, 2016

### Geometry Problem 1234: Quadrilateral, 60 Degrees, Midpoint, Congruence, Cyclic Quadrilateral, Concyclic Points

Labels:
60 degrees,
circle,
concyclic,
congruence,
cyclic quadrilateral,
midpoint,
quadrilateral

Subscribe to:
Post Comments (Atom)

https://goo.gl/photos/xakCyYvXLCEezDqw6

ReplyDeleteAngle B supplement to angle D => qua. ABCB is cyclic

Triangle ABC is equilateral ( 3 angles= 60 degrees)

Let m= AB=BC=AC

We have BF. BG= m/2 . 2m= m^2….. (1)

Triangles ABE similar to DBA ( case AA)

So BE/AB=AB/BD => BE.BD= BA^2= m^2…. (2)

Compare (1) to (2) we have BF.BG=BE.BG= m^2

So Points D,E,F,G are concyclic

BADC is cyclic => ABC equilateral and Tr. BAE ~ BDA (AAA)

ReplyDeleteBE/BA = BA/BD

BA.BA=BE.BD ----------(1)

2.BF.BC=BE.BD

2.BF.BG/2 = BE.BD

BF.BG=BE.BD Q.E.D