Saturday, July 9, 2016

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

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.


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

2 comments:

  1. https://goo.gl/photos/xakCyYvXLCEezDqw6

    Angle 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

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

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