Sunday, April 30, 2017

Geometry Problem 1334. Square, Kite, Sum of Segments

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 1334. Square, Kite, Sum of Segments.

7 comments:

  1. Problem 1334
    Is <GBE=<EBC=<EHG then GH=BG. Draw BK perpendicular in BE at point B (K belongs to ΑD).But triangle KBA=triangle BCE (rectangular, AB=BC,<KBA=<EBC) so KA=CE.Now rectangular triangle KBH is <KBG+<GBH=90=<BKH+<BHK so <KBG=<BKG ,then KG=BG=GH. Therefore GH=KG=KA+AG=CF+AG.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

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  2. ∆EDH ~ ∆BCE => x²= ab + bc ( x = AD )
    From ∆ABG => x² = c² - a²
    From two rows c = a + b

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  3. Extend DA to X such that AX = b.
    Then Tr.s BCE,BEF & AXB are all congruent SAS and so < ABX = < GBH = < CBE < GHB.

    Since AB is perpendicular to XH, it follows that < XBH = 90

    But GB = GH so each of these = XG
    Therefore a+b =c

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  4. Solution 2

    Let AB = p

    From similar triangles b/p = p/(a+c)
    So p^2 = b(a+c) = c^2 - a^2 and hence dividing by a+c, b = c-a

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Draw BK ⊥ BE, K on DA. Draw circle through H, B, K
    G center of the cirle

    ReplyDelete
  6. Triangle GBH is Isosceles and GB=a+b (One way to derive is considering the right triangle BFE and applying pythogorus)
    => c=a+b

    ReplyDelete