Monday, May 19, 2008

Elearn Geometry Problem 49

See complete Problem 49
Angles, Triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. Let alpha =a. From Tr. ABC, we've Ang EAF = 90 - 3a. EF=AEsin(90-3a)=AEcos(3a). But AB/AE=cos(a). Thus EF=ABcos(3a)cos(a). Hence, (AB+EF)/2 =(ABcos(3a)/cos(a)+AB)/2 =(AB/2)(cos(a)+cos(3a)/cos(a)) =(AB/2)(2cos(a)cos(2a))/cos(a)) = ABcos(2a) or (AB+EF)/2 =ABcos(2a)-----(1). Now from Tr. ABD BD/AB =sin(90-2a) =cos(2a) or BD=ABcos(2a)----(2)
    (1) & (2) give us, BD=(AB+EF)/2

  2. It's possible to solve this problem without using trigonometry.
    The key is auxiliary construction:

    Geometry problem solving is one of the most challenging skills for students to learn. When a problem requires auxiliary construction, the difficulty of the problem increases drastically, perhaps because deciding which construction to make is an ill-structured problem. By “construction,” we mean adding geometric figures (points, lines, planes) to a problem figure that wasn’t mentioned as "given."

  3. AB=AE cos alpha and EF=AE cos 3alpha.Then add the two to get the desired result.

  4. Let’a write a for alpha. Take G symmetric of A over the angle C’s bisector, and GH perpendicular to AC.
    We have ang(BAD) = 90º-2a, and ang(GAC) = 90º-a, so ang(BAG) = a. That means that AEG is isosceles, with GB = BE.
    Thus B is middle point of GE, then GH - BD = BD - EF. But GH = AB by symmetry,
    so AB – BD = BD – EF and BD = (AB + EF)/2.

  5. To Newzad
    Please let me know what is the language you write in and the corresponding nationality.
    (Just curiosity. I'm brazilian and speek portugese).

  6. Let AJ be perpendicular to CB, J on BC so that < JAB = 2@. Let AH be the bisector of < JAB, H on BC

    So < JAH = < HAB = < BAE = @. Now drop a perpendicular from H to AC, HG with G on AC

    Tr. s AHG and ABE are congruent ASA since Tr.s ABE and ABH are congruent ASA

    So AB = HG and the result follows since HB = BE

    Sumith Peiris
    Sri Lanka

    Draw BG and GH as per sketch
    Observe that ∠ (GAB)= alpha and ∠ (GAE)= 2.alpha
    Triangle GAE similar to tri. GAC … ( case AA)
    So both tri. GAE and GAC are isosceles
    In isosceles triangle AGC , heights AB and GH are congruent
    In trapezoid GHFE we have BD=1/2(GH+EF)= ½(AB+EF)

  8. Reflect AE along AB to AK, triangle ABE = ABK;
    area AEC = area ABC- area AEK; area AEC =1/2*EF*AC;
    area AKC =area ABC + area ACK =area ABC + area ABE; area AKC=1/2*AB*KC=1/2*AB*AC;
    area AEC+ area AKC = 2 area ABC; area ABC=1/2*BD*AC;
    so BD=1/2(AB+EF);

  9. With the great help of my friend Greg, we have found an original solution.
    See original solution on this video

    ang(BAC) = 90 - 2α => ang(BAD) = 90 - 2α and ang(ABD) = 2α
    Let FJ parallel to BC, J on BD
    BJFE is a parallelogram, BJ = EF and ang(DFJ) = 2α
    ABE right in B and AEF right in F => A, B, E and F are concyclic, the center O of the circle being the middle of AE
    Let BD intersect C in K
    ABO isosceles in O => ang(ABO) = α and ang(ABK) = 2α : BO is the angle bisector of ang(ABK) and AB = BK
    ang(ABK) = ang(AFK) = ang(DFK) = ang(DFJ) = 2α so on BD J and K are mirror images of one another about AC
    So DJ = DK, 2BD = BJ + JD + BK – DK and since BJ = EF and BK = AB then 2BD = EF + AB => BD = (AB+EF )/2