Tuesday, March 21, 2017

Geometry Problem 1324: Quadrilateral, Diagonal, 45 Degrees, Angle Bisector, Isosceles Triangle, Congruence

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.

Details: Click on the figure below..

Geometry Problem 1324: Quadrilateral, Diagonal, 45 Degrees, Angle Bisector, Isosceles Triangle, Congruence.

Posted by Antonio Gutierrez at 4:56 PM
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1 comment:

  1. Peter TranMarch 21, 2017 at 7:30 PM

    https://goo.gl/photos/wKXAUAr4ELQD6xxu8

    Let M is the midpoint of AE
    We have ∠ (CAE)= 45 and AE⊥CD
    Let BC= x => AM=ME= x , BM=CE=2x ; and AC=CD= 2x.sqrt(2)
    Apply Pythagoras’s theorem in triangles BCD and ABM we have
    BD= 3.x and AB= x. sqrt(5)=> CF= x/2. Sqrt(5)
    Apply Pythagoras’s theorem in triangle BCF we have BF^2= 9/4. x^2
    So BF= 3/2.x= BD/2

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