Monday, January 5, 2009

Elearn Geometry Problem 220: Right Triangle, Altitude, Angle Bisector, Distance, Arithmetic Mean

 Geometry Problem 220. Right Triangle, Inradius, Altitude, Angle Bisector, Distance, Arithmetic Mean

See complete Problem 220 at:

Level: High School, SAT Prep, College geometry


  1. AB,BC,CA touch the incircle with centre O at X,Y,Z respectively,
    mBFA=pi/2-mDBF=pi/2-mFBC=mABF, so AB=AF, so r=XB=ZF, similarly r=ZE, then r=EF/2=(a+b)/2

  2. (1)triangle BDC is rt.therefore m angle BCD=2 violet angles.
    (2)by exterior angle theorem m angle AFB = 2 VIOLET PLUS 1 GREEN ANGLE.
    (3) therefore by (1) &(2) triangle ABF IS AN ISOSCELES TRIANLE.
    (4) let the center of that incircle be O.let the incircle touch AB,BC,CA at LMN respectively.
    (5)then tringle LOB is congruent ro tringle NOF by SAA theorem of congruence.
    (6) we note that quadrilateral LOMB is a square.
    (7) by (5) and(6)LB=NF
    (8)similarly by proving BEC isoscles and then triangle BOM congruent to EON,BM cong. to EN
    (9) but BM=LB(sides of square.)
    (10)therefore EN=NF
    (12)N midpoint of EF-proved
    (13) therefore NE=NF=r=a+b/2
    (14) hence the proof!!!

  3. we can prove that BG=BD=BH and GE is perpendicular to AB, BC is perpendicular to AB, so AG/AB=AE/AC, AE=AG.AC/AB=AC(AB-BG)/AB=AC(AB-(AB.BC/AC))/AB=AC-BC simillarly CF=AC-AB hence a+b=AC-AE-CF=AB+BC-AC=2r since tr ABC is right triangle r= AB+BC-AC/2 so a+b/2=r

  4. < AFB = < C + < A/2 = < ABF so AB = AF
    Similarly BC = CE

    So AC = AB + BC -(a+b) hence
    a+b = AB + BC + CA = 2r

    Sumith Peiris
    Sri Lanka