Tuesday, September 25, 2018

Geometry Problem 1388: Triangle 40-60-80 degree, Incenter, Congruence

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Details: Click on the figure below.

Geometry Problem 1388: Triangle 40-60-80 degree, Incenter, Congruence.


  1. Make AD = AB, D on AC;
    so ABD is equilateral triangle;
    so angle DBC=20; angle BDC =120 = angle CIB,
    so triangle BDC equal to triangle BIC;
    so CI=BD=AB=c

  2. Let D be the intersection point of the continuation of IC on AB, E be the intersection of the continuation of BI on AC and let x = IC and y = EC.

    1. tr ADC is similar to tr IEC so let k be the scale factor and CD = ky, AC = kx
    2. Tr CBD is isosceles (40-80-80) so BC = CD = ky
    3. Tr BEC is also isosceles (40-100-40) so EC = BE = y
    4. Tr ABC is similar to AEC (40-60-80) so AB/AC = BE/BC = y/ky = 1/k

    which implies AB = AC * 1/k = kx * 1/k = x or in other words AB = IC = x.

  3. Considering usual triangle notations,
    let BI intersect AC at D and CI intersect BA at E
    We know AD=bc/a+c and DC=ab/a+c (Angle bisetor theorem)
    Observe that BEC is isosceles and EC=a
    Similarly BD is isosceles and BD=DC=ab/a+c --------(1)

    Since BDA similar to CBA (AAA)
    => BD=CB.AB/AC
    => BD=a.c/b-----------(2)
    =>c=b^2/(a+c) ------------(3)

    Observe that DIC is similar to EAC
    =>IC=b.(ab/a+c)/a (substitute values for DC and EC)
    =>IC=b^2/(a+c) ------------(4)

    From (3) & (4) the result follows

  4. https://photos.app.goo.gl/M6AHCEaycVHP1j7v5