Monday, May 19, 2008

Elearn Geometry Problem 47

See complete Problem 47
Angles, Triangle, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. good problem
    we have to draw a perpendicular from from B to AC and E is the foot of the perpendicular .prove triangle BED is a rightangled isoscleles triangle
    hence the result

  2. Draw BE,CF perpendicular to AC & AB extended respectively. Let AD=DC=p. From Tr. ACF we've CF = AC/2 = p while angle FBC =30+15=45; hence BF=p and from Tr. ACF, AF=2p*V3/2 =V3*p. Thus, AB=V3p-p. Now angle ABE=60 and so BE=(V3p-p)/2 and AE=(V3p-p)V3/2 which gives DE = p-(V3p-p)V3/2 =(V3p-p)/2. In other words BE = DE or angle DBE=45 deg. Finally, angle ABE + angle DBE + x + 45 = 180 or 60 + 45 + 45 + x = 180 which yields x = 30 deg.

  3. Prolong AB, then draw a perpendicular from C to AB, being F de foot of this perpendicular.
    Then triangle DFC is euqilateral and BDF is isósceles of base BD.

    Completing angles in the diferents triangles we'll have in BDF that 30 + 2(45+x)=180 <=>2x=60 <=> x=30

    Greetings :D!

  4. Drop a perpendicular CE to AB extended. Then Tr. BCE is right isocoles and BC^2 = 2BE^2 = CD.CA

    Hence BC is tangential to Tr. ABD and so x = 30

    Sumith Peiris
    Sri Lanka

  5. Or let E be the foot of the perpendicular from C to AB and let F be on AE extended such that BE = EF

    Then AD = DC = DE = BE = CE = EF

    So < BDF = 90 = < BCF. Hence BDCF is cyclic and < BFC = 45 = < ADB

    Hence x = 30

  6. Take E reflection of C about AB, tr ACE is equilateral, <CBE=<CDE=90, BDCE is cyclic, hence <CBD=<CED=30, done.