Saturday, March 29, 2014

Geometry Problem 999. Right Triangle, Midpoint, Median, Double Angle, Congruence

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

Click the figure below to see the complete problem 999.

Online Geometry Problem 999. Right Triangle, Midpoint, Median, Double Angle, Congruence

Posted by Antonio Gutierrez at 1:17 PM
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12 comments:

  1. AnonymousApril 6, 2014 at 6:27 AM

    See http://bleaug.free.fr/gogeometry/999.png

    BC perpendicular bisector FEG meets AC in F midpoint of AC, and E midpoint of DC (Thales). By construction ∆BEC and ∆BFC are isosceles. Right angle B implies ∆DEB is isosceles and similar to ∆BAE with acute vertex angle = 2x.

    Let H be the intersection of CD and BF. Since ∆BFC is isosceles, so is ∆BHE. Triangles ∆BAE and ∆BHE are isosceles and share the same base, hence AH bisects angle ∠BAE.

    ∠BAH = ∠BCH implies ∆AHC isosceles. Hence 3x=45° or x=15°.

    bleaug

    ReplyDelete
  2. AnonymousApril 7, 2014 at 5:21 PM

    Please elaborate on why "∠BAH = ∠BCH implies ∆AHC isosceles." Thanks.

    ReplyDelete
    Replies
    1. UnknownMay 22, 2014 at 2:42 PM

      as you AHC question why is isosceley

      Delete
    2. UnknownMay 30, 2014 at 3:27 PM

      why "∠BAH = ∠BCH implies ∆AHC isosceles." Thanks.????????

      Delete
    3. Peter TranJune 3, 2014 at 11:05 AM

      Draw circle diameter AC, center F. B will be on this circle
      Let assume that AHC is not isosceles and AB >BC
      Point H is on radius FB. M and N are the projection of H over AB and AC
      Using elementary geometry we can show that HA > HC and HN> HM
      HN/HC > HM/HA so ∠(HCB)> ∠(HAB)
      So If AB> BC , it doesn’t exist point H on FB so that ∠(HCB)= ∠(HAB) ( both angles are positive)

      Delete
    4. UnknownJune 24, 2014 at 4:26 AM

      sorry Mr peter ((Using elementary geometry we can show that HA > HC and HN> HM
      HN/HC > HM/HA so ∠(HCB)> ∠(HAB)))???????????

      Delete
  3. MATHS AND PHYSICS WORLDApril 12, 2014 at 3:06 PM

    A solution is here: http://mfcosmos.com/archives/15066

    ReplyDelete
  4. MATHS AND PHYSICS WORLDApril 19, 2014 at 8:02 PM

    An equally challenging version of the above exercise (with solution) you can find here http://mfcosmos.com/archives/15173

    ReplyDelete
  5. Sumith PeirisNovember 11, 2015 at 11:12 PM

    Easily Tr. ABE is isoceles. Let AE = AB = q, DE = EC = BE = p and BD = r

    Now BE is a tangent to circle ADE hence
    p^2 = rq ......(1)

    Further AD is a tangent to circle ACE hence
    (q-r)^2 = 2p^2 ....(2)

    From (1) and (2) we have
    q^2 = 4p^2 -r^2 ......(3)

    Now consider h the altitude of Tr. ABE
    h^2 = p^2 -r^2/4 = 4q^2 from .....,(3)

    So h = q/2 hence 2x = 30 and so x = 15

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. Sumith PeirisNovember 12, 2015 at 5:44 AM

    H is the altitude of Tr. ABE drawn from E

    ReplyDelete
  7. Sumith PeirisNovember 12, 2015 at 9:07 AM

    Alternatively from (3) above

    4p^2 - r^2 = q^2

    LHS = BC^2 from right Tr. BCD

    So BC = q and < BAC = 45 and hence 3x = 45 and x = 15

    ReplyDelete

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