Wednesday, January 14, 2009

Elearn Geometry Problem 224: Viviani Theorem, Isosceles Triangle

Altitude, Distance, Point on the extension of the base
In an isosceles triangle ABC (AB = BC), the sum of the distances from any point on the extension of CA to the equal sides is equal to the altitude of the equal sides.


Geometry Problem 224. Viviani Theorem
See complete Problem 224 at:
gogeometry.com/problem/p224_viviani_theorem_isosceles_triangle.htm

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 5:41 AM
Labels: , , , , ,

6 comments:

  1. macsim gabriel florinFebruary 2, 2009 at 1:47 PM

    Area of BCD = Area of ABD + Area of ABC. So, using the fact that AB=BC, we have :
    e*AB/2 = f*AB/2 + h*AB/2.So we conclude that :
    e=f+h, from where : h=e-f

    ReplyDelete
  2. Kalyani MaratheJune 11, 2009 at 1:50 AM

    (1) first, join points D and B.
    (2) using standard notation that-
    area of triangle ABC =[ABC]
    we get the equation-
    [ABC]=[DBC]-[DAB]
    (3) [ABC]=1/2AH.BC=1/2h.BC
    (4) [DBC]=1/2DE.BC=1/2e.BC
    (5) [DAB]=1/2DF.AB=1/2f.AB but,AB=BC hence,it is equal to,1/2f.BC
    (6)by (3),(4),(5)and substituting the values,
    1/2h.BC=1/2e.BC-1/2f.BC=1/2BC[e-f]
    (7)1/2 BC get cancelisd on both the sides and we obtain the equation,
    (8)h=e-f
    (9) hence the proof!!

    ReplyDelete
  3. AnonymousSeptember 26, 2009 at 6:06 AM

    Prof Elie Achkar (Lebanon)
    let BG perp AC then G is midpoint AC
    DF perp BA and DA perp BG then
    angle(ADF) = angle (ABG)
    DA perp BG and DE perp BC then ADE = GBC
    ABG = GBC then ADF = ADE and DA biss(FDE)
    let AI perp DE then DF = DI and AH = IE (rectangle )
    DF + AH = DI +IE +DE
    then f + h =e

    ReplyDelete
  4. c .t . e. oDecember 14, 2009 at 10:07 AM

    tr DEC and AHC are similar ( common angle C )
    e/h = (DA+AC)/AC than e/h = DA/AC + 1 (1)

    tr DAF and ACH are similar ( an DAF = C )
    DA/AC = f/h (2)

    substitute (2) in (1)
    e/h = f/h +1 => e/h = f/h + h/h => e = f+h or e-f = h

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  5. PravinJune 13, 2012 at 9:57 PM

    Complete rectangle DEHG
    Rt ∆ADG ≅ Rt ∆AFG
    since ∠ADG = ∠ACB (DG∥AC, alternate ∠s)& they have common hypotenuse AD
    So AG = DF
    Hence e - f = DE - DF = GH - GA = AH = h

    ReplyDelete
  6. ArifJanuary 11, 2018 at 8:16 AM

    Triangle DEC and DFA are similar therefore :

    e/DC = f/DA
    f/e = DA/DC

    Triangle DEC and AHC are similar therefore :

    e/DC = f/AC
    f/e = AC/DC

    Adding these two equations and using the fact that AC+DA =DC

    (h+f)/e= (AC+DA)/DC
    (h+f)/e= 1
    h+f=e
    h=e-f

    ReplyDelete

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