Sunday, May 18, 2008

Elearn Geometry Problem 9

Geometry Problem, Triangle, Angles

See complete Problem 9 at:

Triangle, Angles, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.




  3. I am brazilian, sorry for the bad english.
    Let E be the intersection of AC and BD. For triangle AED, angle(AEB) = x + 60. For triangle EBC, angle(ECB) = angle(AEB) – angle(EBC) = x + 60 – (x + 30) = 30.
    So angle(ADB) = 2.angle(ACB), points A, B and C belong to a same circle with center D, AD = DC, triangle ADC is isosceles and x = 10.

  4. The solution is uploaded to the following link

  5. As Nilton has pointed out: angle(ADB) = 2* angle(ACB) AND AD = BD which implies that D lies on the perpendicular bisector of AB which, in turn, means that D is the circumcentre of Tr.ABC; hence etc.

  6. To Ajit: You're right, the condition AD = BD should be mentioned explicitly, though it is said in the enunciate. Thanks for your help.

  7. Let perpendıcular bısector of AB (passing through D) intersect AC at E; <ADE=30, so <CED=30+x=<CBD and BEDC is cyclic, so <DBE=<DCE=10. As, by symmetry <DAE=<DBE, our answer is <DAE=10.

    Best regards,
    Stan Fulger

  8. Triangle ABD is equilateral; let the perpendicular bisector of AB intersect AB at E and, clearly <CED=30+x=<CBD, so BEDC is cyclic and <DBE=<DCE=10 ( 1 ). Since by symmetry <DAE=<DBE, we have our answer, x=10.

    Best regards,
    Stan Fulger

  9. Extend BD to E such that BD = DE

    Then < EAC = < EBC = x+30 and so ABCE is cyclic with D as centre
    Hence Tr. ADC is isoceles and x=10

    Sumith Peiris
    Sri Lanka

  10. This comment has been removed by the author.


  12. Construct a line AE such that AE = BC and is at an angle of 30 degree to AD.
    Join DE. The trianlges BAE and ABC are congruent (SAS)
    => m(AEB) = m(ACB) =y (say) => A,E,C,B are concyclic.
    => m(AEC) + m(ABC) = 180
    => m(AEB) + m(BEC) + m(ABE) + m(EBC) = 180
    => y+(60-x)+60+(x+30) = 180
    => y = 30
    => D is the center of the circle passing through A,E,C and B
    Now consider the isosceles triangle BDC in which BD = BC
    => x+30 = 40
    => x = 10 degrees

  13. Construct point E such that DE is the perpendicular bisector of triangle BCD, such that <BED=90. Construct point F, which is the intersection point of ED and AC. Join BF.

    Triangle EFB is congruent to EFC (SAS). Since <BCA=30, <EBF=30.
    Hence <FBD=x. Also, BF=CF.
    Since BF=CF, FD=FD, <DFB=<DFC,
    triangle DFB is congruent to DFC (SAS).
    Hence <FBD=FCD,