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

12 comments:

  1. 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

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

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    Replies
    1. as you AHC question why is isosceley

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

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    3. 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)

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

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

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

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  5. 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

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  6. H is the altitude of Tr. ABE drawn from E

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  7. 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

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