Wednesday, May 11, 2016

Geometry Problem 1212: Equilateral Triangle, Equilateral Hexagon, Concurrent Lines

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

This entry contributed by Peter Tran.

Click the figure below to view more details of problem 1212.

Geometry Problem 1212: Equilateral Triangle, Equilateral Hexagon, Concurrent Lines


  1. What is given regarding the hexagon is that the 6 sides are equal - that's all

    Is this the case Antonio / Peter?

  2. Till the concurrence is proved P may be mentioned as the point of intersection of any two of the three segments and the 3rd segment may be shown as two broken, dotted segments.Better still there is no mention of P in the figure.

    1. My drawing with Cabri II reveals the following:
      (i) Triangles ADM, CHG, BFE are congruent to one another.
      (ii) If O is taken as the center of triangle ABC,
      Triangle triplets ODE, OFG, OHM are pairwise congruent;
      Triangle triplets OEF, OGH, OMD are pairwise congruent.
      This infomation might be helpful to co-solvers.

  3. Draw segment DJ equal to MH and parallel to AC such that Angle EDJ =60.Join JH,EJ and GJ.
    DJHM is parallelogram,and JH parallel and equal to DM. Tr.EDJ is equilateral triangle, we have EJ parallel to FG and equal to it. We have parallelogram EFGJ, GJ is equal to EF and parallel to it. It means in triangle GHJ, GH=HJ=JG and it is equilateral. Also MD, EF and GH make angle of 60 deg. with each other.
    It is easy to see that Tr.ADM, Tr. BFE and Tr.CHG are congruent to each other.
    We have AM=BE=CG and AD=BF=CH. Consider point P' as centroid of triangle ABC.We have P'M=P'E=P'G and P'D=P'F=P'H. Also all triangles formed by joining above segments are congruent and each angle formed at P' is 60 deg. Its easy to see that DP'G, MP'F and EP'H are collinear. Hence DG,FM and EH are concurrent at centroid of Triangle ABC.

  4. Kindly elaborate how MD, EF and GH make 60 deg with each other

    1. Triangle GHJ is equilateral
      HJ is equal and parallel to MD
      GJ is equal and parallel to EF
      Hence MD, EF and GH make 60 deg angle with each other.


  5. Let the sides of the hexagon each be = p

    Construct rhombus EFQD from which Tr. FGQ is easily seen to be equilateral.
    So QG//MH, QGHM is a rhombus too and so Tr. DQM is equilateral

    Hence QD = QM = QG = QF, so D, F, G, M lie on a circle with Q as the centre
    and radius = p

    Similarly we can show that D,E,G,H lie on a circle with radius p and
    centre Q.

    So the hexagon DEFGHM is cyclic each side subtending an angle = 60 at Q
    the centre and hence EQM, DQG, FQM are all straight lines meeting at Q

    Sumith Peiris
    Sri Lanka

  6. Extend EF, GH, DM, so that EF intersects with GH at N, GH intersects DM at O and DM intersects EF at P. Also, make circles with radius EF centered on E, G, H, D. They all intersect at a single point called K. This makes equilateral triangles EKD and GKH. Thus, GH = KH = HD = DM. Consequently, DKHM is a rhombus. From the fact that rhombuses have parallel sides, it is deduced that triangle ADM, DMK, KHM and HOM are all congruent triangles. Thus, angle DAM and angle HOM are equal (both 60 degrees as ABC is equilateral). Using similar techniques, it is deduced that OPN has three 60 degree angles. From here, we prove the symmetries of DFGHMD. For example, angle EFB = angle NFG (vertically opposite angles) and angle EBF = angle FNG. Thus, angle BEF = angle FGN. Using similar techniques, it is proved that angle DEF = angle FGH = angle DMH and angle MDE = angle EFG = angle GHM. This and the fact that EFGHMD is an equilateral hexagon show that MF, EH and DG are axes of symmetry of the hexagon. Thus: MF, EH, DG are bisectors of all six angles of the hexagon. Let EH and FM intersect at X, and let GD and FM intersect at Y. Angle XMH = angle YMD and angle XHM = angle YDM, and finally MH = MD, thus YMD is congruent to XMH. Because there is no angle between angle XMH and YMD (as angle DMH is bisected by MF), Y must equal X. QED.


  7. Please explain why "circles with radius EF centered on E, G, H, D. They all intersect at a single point called K."

  8. Kylie Roup and Caitlyn SolomonJanuary 17, 2018 at 11:15 AM

    Using Geogebra a line was rotated 60 degrees and connected to another line, creating a 60 degree angle, which was angle B. Then a random point on both of the lines was chosen to become the third side of the small triangle BEF, making line EF. A circle was made using E as the center. Using EF as the radius it made ED another radius of the circle. Another circle was made using F as the center making FG a radius. Because they are all radii, the lines are equal. A line was made across D and G. This gives you a quadrilateral EFGD. Then the quadrilateral was reflected across line DG to make an equilateral hexagon. Line AC was made and we made sure triangle ABC was equilateral by checking that the angles A, B, and C were all 60 degrees. Then quadrilateral EFGH was reflected across line EH to make sure it was an equilateral hexagon. Because quadrilateral EFGH was reflected across line EH, the angles DEF and DMH must be equal. Line DE is equal to DM and EF is equal to MH because they are sides of an equilateral hexagon. Triangle DEF is equal to DMH because of side-angle-side. Using the same reasoning, triangle FGH is also equal to the other triangles. According to Ceva’s theorem, if ED/EF multiplied by GF/GH Multiplied by MH/MD is equal to one, then the lines are concurrent. Because of CPCTC, they are equal, so they are also equal to one. Therefore, lines DG, EH, and FM are concurrent.

    (Ceva’s theorem found on

  9. How to construct DE on AB, FG on BC and MH on AC since a line free and hold the drag