Thursday, March 12, 2020

Geometry Problem 1461: Quadrilateral, Triangle, Angles, 30-60 Degree, Congruence, Auxiliary Lines

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

Details: Click on the figure below.

Geometry Problem 1461: Quadrilateral, Triangle, Angles, 30-60 Degree, Congruence, Auxiliary Lines, iPad apps, Tutoring.


  1. Draw a perpendicular from B to AC to meet AC at X
    Draw a perpendicular from C to AD to meet AD at Y

    Now < BCA = 35 and so AB = BC and AX = XC = CY since Tr. ACY is 30-60-90.

    Therefore Tr.s BXC and CYD are congruent ASA

    So BC = CD and x = 25

    Sumith Peiris
    Sri Lanka

    1. @Sumit Peiris your thinking ability is great to tackle any geometry problem. Buddy will u plz tell me how did you practice in starting and how can you able to solve this type of problem with simple and elegant method.. your soln of each and every problems are awesome and mind blowing. Will u plz give me some suggestions?

    2. Dear Manish - Thank u for your kind words. Whatever small ability I have I give glory to Jesus Christ my Lord. I'm now 56 but I loved Geometry from the age of 15.

      Some suggestions - draw a neat diagram or print, try to see where known theorems fit. See if any construction is needed. Work backwards when necessary.

      These days WHO says TEST TEST TEST. For Geometry PRACTICE PRACTICE PRACTICE!!!And no better site than Antonio's Gogeometry site.

      I must add that I am an admirer of the solutions provided by Gogeometers like
      Peter Tran, Pradyumna Agashe, Apostolis, Stan Fulger, Praveen and many others.

      Trying my best to solve Problem 1470 these days of curfew in Sri Lanka but no luck yet!

      Cheers Manish!

  2. Let E be the intersectio of AB and CD; <AED=60 degs. Take F reflection of D across AC; tr. ADF is equilateral and as <AFC=<ACD=55 degs, B is the circumcenter of tr. ACF, hence BF=BC=AB thus BD is perpendicular bisector of AF, making <ADB=30 degs, i.e. <BDC=25 degs. Or, toextend the proof little bit, notice that the intersection of AC and BD is the circumcenter of tr. ADE, etc.

    Best regards

    1. how did BF=BC=AB came ? i haven't seen why ?

    2. Because, as I have proved, B is the circumcenter of triangle ACF.

  3. Let triangle CDE=equilateral.Is E=circumcenter of triangle ACD,AB=BC=CE=AE=DC.Therefore <CDB=<CBD=25.

  4. < BCA = 35 hence AB = BC and CXY is equilateral so AX = XC = XY = CY

  5. Dear sumith Peiris
    May you help me for solution of another problem ,that is problem 199 ,that is very kind of you

    1. No, I haven't been able to solve it yet

    2. Hello Unknown:

      Problem 199 was solved by Mixalis Tsourakakis in 2013, in a solution posted by Antonio accessible through a link.
      The posted solution was in Greek. I have made a translation of it (using Google translate so it is imperfect but still understandable) which reads as follows:

      We easily find that ∠DAB = 52°, ∠ECA = 8°. We consider the perpendicular from B to AE. Then GE orthogonal to AB is the third height of the triangle ΑΒG.

      We consider L symmetrical of A to D and let Z be the intersection of MG with AL and we get ZC. ∠BLA=52°, ∠DBL=38° and ∠MGA = ∠LGZ = 38°.

      It is ∠GEC + 8° => ∠MGA = 38° and so ∠ZEC = 30°. Still ∠ZBC + ∠BCL = 52° = ∠BLA and since ∠BCL = 22° then ∠ZBC = 30° so the quadrilateral BCZE is cyclic and ∠ΒΖE = ∠ΒCE = 14°.

      We consider the perpendicular GK to BZ. Since ∠BLA = 52° then ∠LGK = 38° from ∠LGZ = 38° GC is a bisector of ∠ΚGZ. If the perpendicular from Z to ΑC intersects GK in Η, since ∠GCZ = 30° (because ∠CGZ = ∠MGA = ∠ABD = 38° and ∠EZC = 180° - ∠CBD = 112°) the triangle HCZ will be equilateral so ∠HZC = 60° because ∠ΚΙΒ = ∠HΙC = 60° the quadrilateral ΙHCZ is inscribed in a circle so ∠GIZ = 60° and so in the triangle ΒΙΖ IK is both the height and the bisector and therefore also the median of BZ.
      But the triangle BGZ is isosceles and therefore ∠GZB = ∠ΒΖG = 14°.
      So ∠ΒGM = 28° but since ∠ΒGM = ∠ΒΑΝ then ∠ΒΑΝ = 28° and therefore χ = 52°-28° <=> χ = 24°.


      I found it quite brilliant!

    3. Problem 199

      Michael from Greece has given a good solution (in Greek) and I use his diagram in my slightly modified proof below
      Construct isosceles Tr. ABL, L on AC. Drop a perpendicular EM from E to AB. Let ME extended meet DL at G & BL extended at Z. So E is the orthocenter of Tr. ABG, AEN is perpendicular to BG, N on BG.
      Now complete kite CZGH with diagonals perpendicular meeting at X. Let BZ, GH meet at K and let GH, BC meet at I.
      Easily < HGC = 38, < GLK = 52 so < GKL =90. < CEZ = CBZ = 30 so BCZE is concyclic and hence < EZC = < EBZ = 30 so that Tr. HCZ is equilateral which makes CHIZ concyclic since < CHZ = < CIZ = 60
      Therefore < HCI = 8 = < HZI so < BZI = 30 = < IBZ which makes BIZG a kite with < BZG = < GBZ= 14
      Hence < DBG = 38-14 = 24 = x since ABND is concyclic

  6. How did CXY is equilateral?

  7. My solution at