Wednesday, January 21, 2015

Geometry Problem 1076: Square, 45 Degrees, Angle, Sum, Distance, Auxiliary Construction

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1076: Square, 45 Degrees, Angle, Sum, Distance, Auxiliary Construction.

4 comments:

  1. Draw circle centered A, radius=AD
    Locate point H on the circle such that ∠ (HAF)= ∠ (DAF)= alpha
    Triangles FAD congruence to HAF ( Case SAS) => FH=FD= b and ∠(AHF)=90
    We have ∠ (EAH)=45- alpha=∠ (BAE) => triangles EAH congruence to EAB
    So EH=EB=a and ∠ (EHA) 90
    So x= a+b

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  2. Rotate anti-clockwise 90°, center at A.
    Then D→B. Let F→G.

    Then AF=AG and ∠EAF=∠EAG.
    So ΔEAF congruent to ΔEAG.

    Hence, x=EF=EG=a+b.

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  3. Fold AB, AD along AE, AF respectively.
    B, D land at the same point P on EF.
    So x = EP + PF = EB + DF = a + b.

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  4. Extend CD to G such that DG = a, then Tr.s ABE and ADG are congruent from which we can deduce that Tr.s AFE and AGF are also congruent.
    Hence x= FG = a+b

    Sumith Peiris
    Moratuwa
    Sri Lanka

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