## Tuesday, February 21, 2023

### Geometry Problem 1509: Congruence of Triangles in a Trapezoid and a Square, Measurement. Difficulty Level: High School

Geometry Problem 1509. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below. More Details

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1. Extend FB to point G such that BF=BG. Then we get CF=CG=CD, hence C is center of circle passing through points D,F and G. Since Angle DCF=90, Angle DGF=90/2=45.

1. Very elegant solution Pradyumna, much better than my own

2. Thank you Sumith for your kind comment.

2. Extend CF to G
Let the side of the square be a, let GF = t and AG = q

From similar triangles BCF, AGF and GCD
5/a = 2/t = a/(q+x).....(1)

Since AFCD is concyclic, t(a+t) = q(q+x)....(2)
Substituting from (1) (2a/5)(a+2a/5) = (a^2/ 5 - x)(a^2/ 5)
14/5 = a^2/ 5 - x and a^2 = 14 + 5x...(3)

But x^2 +2^2 = 2.a^2
Therefore x^2 + 4 = 28 + 10x from (3)
x^2 - 10x - 24 = 0
(x+2)(x-12) = 0

So x = 12 since x = -2 is not permissible

Sumith Peiris
Moratuwa
Sri Lanka

3. https://photos.app.goo.gl/tM61SHQ3HgmUEZrp9

Let H is the project of C over AD
Note that triangle FBC congruence to DHC ( case AA)
So HD=BF=5 and CH=AB=7

1. Good solution, Peter

2. Thank Sumith

4. c.t.e.o.https://photos.app.goo.gl/4E2MnjC9GN6Xd7fz9

5. https://photos.app.goo.gl/4E2MnjC9GN6Xd7fz9