Harvey Littleman

From Gogeometry Using Additional points congruent triangles Equilateral triangle Area of a triangle •ABCDEF is a regular hexagon with center O and side c => AB//DE, BC//EF et CD//FA •Let R=OA=OB=OC=OD=OE=OF •Define G such as PG⊥ AB. The same for H,I,J,K,L •AOF is an equilateral triangle with height h, such […]

Gogeometry Problem 983

 dans Euclide Geometry Problem par
From Gogeometry Using Pythagoras congruent triangles Similar triangles Equilateral triangle Area of a triangle Tangents to a circle Inradius in right triangle •Pythagoras on triangle ΔBFC => (1) BF^2+CF^2=BC^2=a^2 •Incircle in a right triangle => (2) 2x=BF+CF-BC =BF+CF-a •[ABCD]= 4 triangles areas + square with side 2x => (3) S=a^2=2BF.CF+4x^2 […]

Gogeometry Problem 491

 dans Euclide Geometry Problem par
From Gogeometry Using congruent triangles Similar triangles •AD altitude => ∠BDA = π/2 •BE angle bisector =>∠ABE=∠CBE= π/6 •Define M as the intersection of BE and AD •Define F on AB such as BF ⊥ FM •Then we have to demonstrate that FB=FA to prove that FG is perpendicular bisector […]

Gogeometry Problem 663

 dans Euclide Geometry Problem par
From Gogeometry Using central and inscribed angle congruent triangles Concyclic points Tangents to a circle Inscribed angle and tangent to a circle Inradius in right triangle •Point I is on the intersection of the internal angle bisector•=> AI and IB are internal angle bisector •Point O is on the intersection […]

Gogeometry Problem 623

 dans Euclide Geometry Problem par
Gogeometry Problem 315 From Gogeometry Using Similar triangles Define n=CDΔBDB’ is similar to ΔCDC’ => BD/BB’=CD/CC’=>(b+c+n)/b=n/c => b+c+n=bn/c => n(b/c-1)=b+c=> (1)  n=c(b+c)/(b-c)ΔADA’ is similar to ΔCDC’ => AD/AA’=CD/CC’=>(a+2b+c+n)/a=n/c => ac+2bc+c^2+nc=an => n(a-c)= ac+2bc+c^2=> n= (ac+2bc+c2)/(a-c) => n= c(a+2b+c)/(a-c) (2)(1)&(2) : (b+c)/(b-c)= (a+2b+c)/(a-c)=>(b+c)(a-c)=(a+2b+c)(b-c)=>ab-bc+ac-c^2=ab-ac+2b^2-2bc+bc-c^2=> 2ac=-2c^2=>ac=c^2

Gogeometry Problem 315

 dans Non classé par
From Gogeometry Using Area of a triangle Tangents to a circle Inscribed circle of a triangle b=ACAN=AM, AF=AE, MF=NE,CP=CO, CF=CG, PF=OG,BE=BGr=DF=DE=DGx=JM=JN=LO=LPMP=2x S=[ABC]=r(AB+BC+AC)/2 => S=r(AB+CB+b)/2 AB=AE+EB=AF+EBCB=CG+GB=CF+GBAB+CB=AF+CF+EB+GB=b+2BE=> (1) S=r(b+BE) But S=[AMJN]+[MFDENJ]+[PFDGOLP]+[EBGD]S=xAM+(r+x)MF+xCP+(r+x)PF + rEB, knowing that [EBGD]=(rEB+rBG)/2=rEBS=x(AM+MF+CP+PF)+r(MF+PF) + rEB(2) S=xb+2xr + rEB (1) and (2) => S=r(b+BE)=xb+2xr+ rEBbr=xb+ 2xr Therefore x=br/(b+2r)

Gogeometry Problem 418

 dans Euclide Geometry Problem / Géométrie Euclidienne  étiqueté Equal Tangent circles / Incircle / Inradius / radius / Triangle par
From Gogeometry Using Inscribed angles in a circle Equilateral triangle Concyclic points ∠BAD= π/2 => B, A and D are concyclic, circle with center F middle of BD=> FB=FA=FD and ∠BAF=15°In the same way, GC=GA=GE and ∠CAG=15°=>∠FAG=90-15-15=60°ΔFAG is isosceles in A (AF=AG) and ∠FAG=60° => ΔFAG is equilateral Therefore x=FG=BD/2=CE/2=1

Gogeometry Problem 406

 dans Euclide Geometry Problem / Géométrie Euclidienne  étiqueté 15 degrees / Congruence / Midpoints / Right Triangle par
Definition : In euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. Property : Every kite is orthodiagonal, meaning that its two diagonals are at right angles to each other

Kite

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