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.

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## Tuesday, March 21, 2017

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

Labels:
45 degrees,
angle bisector,
congruence,
diagonal,
isosceles,
quadrilateral,
triangle

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https://goo.gl/photos/wKXAUAr4ELQD6xxu8

ReplyDeleteLet 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