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.

Posted by Antonio Gutierrez at 11:51 AM
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12 comments:

  1. APOSTOLIS MANOLOUDISApril 30, 2017 at 12:47 PM

    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

    ReplyDelete
  2. c.t.e.oApril 30, 2017 at 3:39 PM

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

    ReplyDelete
    Replies
    1. KAMAL SINGHMay 20, 2018 at 10:09 PM

      Excellent solution

      Delete
    2. AnonymousDecember 25, 2018 at 11:41 PM

      Can you please explain the first line?

      Delete
    3. c.t.e.oDecember 26, 2018 at 9:02 AM

      EC/BC=DE/DH, EC=b, BC=x, DE=x-b, DH=c-(x-a)

      Delete
  3. Sumith PeirisApril 30, 2017 at 8:59 PM

    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

    ReplyDelete
  4. Sumith PeirisApril 30, 2017 at 9:03 PM

    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. Sumith PeirisApril 30, 2017 at 9:10 PM

    Further result < AFC = 135

    ReplyDelete
  6. c.t.e.oMay 1, 2017 at 1:07 PM

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

    ReplyDelete
  7. AnonymousMay 1, 2017 at 5:49 PM

    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
  8. AnonymousJune 21, 2017 at 4:21 AM

    Let length of square be 1.

    <CBE=<EHD
    <CBE= tan(-1)b
    <EHD= tan(-1){1÷(a+c)}
    So, b(a+c)=1

    BG=GH
    SQRT(1+a^2)=c (∆BAG)
    By substituting b(a+c)=1 into SQRT(1+a^2)=c, you get a+b=c.

    ReplyDelete
  9. Sailendra TApril 28, 2021 at 6:35 PM

    Read @ as Alpha
    Let m(CBE)=m(GBE)=@
    It can be observed that BGH is isosceles=> BG=c
    Mark O on GB such that GO=a
    It can be derived that m(BAO)=@. Extend AO to meet BC at P and BE at Q
    Since m(BAQ)=@ and m(ABQ)=90-@=>BQ _|_ AQ and hence BOP is isosceles => BP=BO
    Tr. ABP congruent to BCE (ASA) => BP=CE=b => BO=b
    GO+BO=a+b=c

    ReplyDelete

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