Wednesday, April 8, 2009

Problem 25: Right triangle, Altitude, Incircles, Incenter, Angle Bisector

Proposed Problem

Problem 25: Right triangle, Altitude, Incircles, Incenter, Angle Bisector.

See complete Problem 25 at:

Level: High School, SAT Prep, College geometry


  1. Let incircles of Tr ABD & Tr.CBD meet AB & BC respectively in J & K. Now with the standard nomenclature we can show that: BD=ac/b,AD=c^2/b
    Hence, r1=Inradius of Tr.ABD=(ac/b*c^2/b)/(ca/b+c^2/b+c) = ac^2/(b(a+b+c))
    Likewise, r2=Inradius of Tr.CBD=(ac/b*c^2/b)/(ca/b+c^2/b+c) = ca^2/(b(a+b+c))
    Now take BA & Bc as the x-axis and y-axis. It can be shown by elementary geometry that F is
    [ac^3/(b(b-a)(a+b+c)), ac^2/b(a+b+c)] since BJ is r1*cot(B/2). Likewise, G is [ca^2/b(a+b+c),ca^3/(b(b-c)(a+b+c))]. By the distance formula which is nothing but using Pythagoras: FG^2 = (ac)^2*(a^+c^2-ab)^2/(b(a-b)(a+b+c))^2 +ac)^2*(a^+c^2-bc)^2/(b(b-c)(a+b+c))^2
    Now a^+c^2=b^2; thus, FG^2 = 2(ac/(a+b+c))^2
    But r = Inradius of Tr. ABC = ac/(a+b+c)
    Hence, FG^2 =2r^2 ---(1) while from Tr. BLE we've BE^2=BL^2+LE^2 where L is the point at which incircle of Tr. ABC touches AB. Also BL=LE=r. Thus, BE^2 =2r^2 ---(2)
    By equations (1) & (2) we've, FG=BE=V2*r
    Now slope FG = [ca^3/(b(b-c)(a+b+c)) - ca^2/b(a+b+c)]/[ca^2/b(a+b+c)-ac^3/(b(b-a)(a+b+c))]
    = ac/(a+b+c)/ -[ac/(a+b+c)]= - 1 which means FG makes an angle of 135 deg with x-axis BA and since BE is the angle bisector of angle ABC (=90 deg) it makes an angle of 45 deg to the x-axis i.e. to BA which in turn means that BE is perpendicular to FG.

  2. Let r, r1, and r2, are the inradii of the triangles ABC, ADB, and BDC respectively.
    Then r=(AB*BC)/(AB+BC+CA),r1=(AD*BD)/(AB+BD+DA) and r2=(CD*BD)/(BC+CD+DB). Hence we get
    (1) r^2 =r1^2+ r2^2.

    (2) Since BE=sqrt(2) *r and
    FG^2 =(r2-r1)^2+(r2+r1)^2 =2(r1^2 +r2^2),
    we get BE=FG.

    (3) Let X and Y be points on BA and BC, respectively.
    Since two triangles ABD and DC are similar,
    FD/XF=GD/YG, that is, XY // FG.
    Now, we see that XY is perpendicular to BE and hence FG is perpendicular to BE.

  3. To Bae deo rak

    It is not clear to me how do you get r^2 =r1^2+ r2^2 from r=(AB*BC)/(AB+BC+CA),r1=(AD*BD)/(AB+BD+DA) and r2=(CD*BD)/(BC+CD+DB)
    Please explain

  4. To Bae Deok Rak
    In (3), what are point X, Y? Without defining their positions, it is impossible to understand the proof.
    Please answer my question.

    Hope that my comment below will answer to questions of Peter and Nilton Lapa
    Let r, r1, r2 are radii of incircles of triangles ABC, ADB and BDC ( see sketch)
    1. Since these triangles are similar so r1= r. (AB/AC) and r2=r.(BC/AC)
    We have r1^2+r2^2= r^2.(AB^2+BC^2)/AC^2
    But AC^2=AB^2+BC^2 so r1^2+r2^2= r^2
    BE^2= 2*r^2 and FG^2 =(r2-r1)^2+(r2+r1)^2 =2(r1^2 +r2^2)= 2.r^2
    So BE=FG

    2. Let alpha= ∠ (ACB)= ∠ (ABD) and x= ∠ (KGH), y=∠ (KBE)
    Since ADB similar to BDC so r1/r2=AB/BC= tan(alpha)
    We have tan(x)=(r2-r1)/(r2+r1)= (1-tan(alpha))/(1+tan(alpha))
    Tan(y)= tan( m(ABE)-m(ABD))= tan(45-alpha)= (1-tan(alpha))/(1+tan(alpha))
    So x=y
    Since ∠ (KHG) complement to x so ∠ (KHG) complement to y .
    In triangle BHL, ∠ (KHG) complement to ∠ (KBE) so BE perpendicular to FG

  6. I had the idea below, I’d like to know your opinion.
    Let L on AB, and M on BC, be the tangent points of the incircle of ABC. Let P on AB be the tangent point of the incircle of ABD, and let Q on BC be the tangent point of the incircle of BDC. BLEM is a square with side r, which diagonals LM and BE are congruent and perpendicular. From problem 023, BD=r+r1+r2. As LP=BP-BL, BP=BD-r1, and BD=r+r1+r2 (this is the result of Problem 023), then LP=r2. Similarly, MQ=r1. So, LPF and GQM are congruent, with LF=MG. As LP and QG are paralel, and PF and MQ are paralel too, then LF and MG are paralel. This shows that LMGF is a paralelogram, with LM and FG congruent and paralel. But LM and BE are perpendicular, so BE and FG are congruent and perpendicular.

  7. FG^2 = (r1-r2)^2 + (r1+r2)^2 = 2(r1^2+r2^2) = 2r^2 by referring to my proof for Problem 24

    But BE^2 = 2r^2 by considering the square whose diagonal is BE

    Hence BE = FG

    For the 2nd part of this problem I salute Nilton Lapa's excellent solution above.

    Sumith Peiris
    Sri Lanka

  8. FG//EL and FG=EL
    See my solution at

  9. The first part:

    From the point E draw a line perpendicular onto the line BD. The line meets at the point Q. In problem 23 we have BD= r+r1+r2, therefore :


    and from problem 28 we have

    EQ= r2-r1

    Draw a line from the point G perpendicular onto AC. The line meets at the point N.
    Then draw a line from the point F perpendicular onto GN. The line meets at the point M.

    Then we have :



    GM = r2-r1

    The triangles FMG and BQE are congruent from SAS and FG=BE

    Second part :

    We have :

    ∠GFM =∠EBQ=u

    Let the line FG meet the line BD at the point P, then :


    If we extend the line BE to meet FG at a point K and let :


    than the angles of the triangle of BKP :