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