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.

See complete Problem 224 at:

gogeometry.com/problem/p224_viviani_theorem_isosceles_triangle.htm

Level: High School, SAT Prep, College geometry

## Wednesday, January 14, 2009

### Elearn Geometry Problem 224: Viviani Theorem, Isosceles Triangle

Labels:
altitude,
distance,
isosceles,
perpendicular,
triangle,
Viviani theorem

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Area of BCD = Area of ABD + Area of ABC. So, using the fact that AB=BC, we have :

ReplyDeletee*AB/2 = f*AB/2 + h*AB/2.So we conclude that :

e=f+h, from where : h=e-f

(1) first, join points D and B.

ReplyDelete(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!!

Prof Elie Achkar (Lebanon)

ReplyDeletelet 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

tr DEC and AHC are similar ( common angle C )

ReplyDeletee/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

Complete rectangle DEHG

ReplyDeleteRt ∆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