Saturday, February 21, 2009

Problem 19: Isosceles Right Triangle, Excenter, Perpendicular, Congruence

Proposed Problem
Try to use elementary geometry (Euclid's Elements)

Problem 19: Isosceles Right Triangle, Excenter, Perpendicular, Congruence.

See complete Problem 19 at:
gogeometry.com/problem/problem019.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

6 comments:

  1. Let B:(0,0), A:(0,a) and C:(a,0)
    Area ABC=a^2/2, AC=V2*a where V=square root
    Hence exradius =(a^2/2)/(2a -v2a) = a/(2-v2)
    Hence AD^2 = (a/(2-v2))^2 + (a/v2)^2
    which gives AD ~ 1.8477590650*a
    while BE=a*cos(ABE)=acos(22.5)~ 0.923879532*a
    2BE = 2*0.923879532*a = 1.847759065*a
    Hence AD = 2BE
    Ajit: ajitathle@gmail.com

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  2. from triangle BEA BE=AE*tan67.5, from triangle BED BE=(AE+AD)*tan22.5,then
    AE=BEcot67.5 and AE=BE*cot22.5-AD. By using last two relations we get BE*cot67.5=BE*cot22.5-AD, then BE(tan67.5-tan22.5)=AD, BE(sin(67.5-22.5)\(cos67.5*cos22.5))=AD , BE(sin45\(sin22.5*cos22.5))=AD then AD=2BE
    Shahlar Maharramov, Yasar University ,Izmir,Turkey and Cybernetics Institute Azerbaijan

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  3. http://img534.imageshack.us/img534/2876/problem19.png
    BE cut DF at G.
    Note that BFDH is a square >> BF=DF ( see picture)
    Triangle BFG congruence to Tri. DFA (Case SAS)
    So AD=BG
    In Triangle GDB , ED is an angle bisector and ED is an altitude so BGD is isosceles
    So BG=2BE=AD
    Peter Tran

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  4. http://henrik-geometry.webs.com/19.htm

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  5. Complete the square BFDG and let DF & BE meet at H. Then Tr.s HFB & AFD are congruent SAA. So the 2 hypotenuses are equal. BE = EH since DE is both the altitude and the angle bisector

    So 2BE = AD

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. my solution here:
    https://www.facebook.com/photo.php?fbid=10206216363756695&set=a.10205987640598759.1073741831.1492805539&type=3&theater

    ReplyDelete