From Gogeometry Using Additional points Inscribed angles in a circle congruent triangles Concyclic points

Gogeometry Problem 412



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


Using Additional points congruent triangles Area of a triangle [OQL]=OL . KL/2=bc/2[OPQ]+[LKQ]=PQ . KL/2+QK.KL/2=(PQ+QK).KL/2=PK.KL=bc/2=>[OQL]=[OPQ]+[LKQ]=bc/2[OPKL]= [OQL]+[OPQ]+[LKQ]=2bc/2=bc In the same way [LMHI]= [IJEF]= [FGNO]= [OPKL]= bc[QRST]= [OLIF] + [OQL] + [LRI] + [ISF] + [FTO][QRST]= c^2+2bc=c(c+2b) Therefore [QRST]= ac (Situation A) But, this is not true if b is too big, then […]

HP008


From Gogeometry Define a=AB=BC=AC DC and DB tangents to Circle O1 => DG=DE=bAD and AB tangents to Circle O1 => AG=AJ=dDB and BA tangents to Circle O1 => BE=BJ=c DB and BC tangents to Circle O2 => BF=BK=eDC and BC tangents to Circle O2 => CI=CK=f a=d+c=e+fDF=b+c+e, DI=b+d+a+fDC and DB […]

Gogeometry Problem 1372



From Gogeometry Using : Additional points Inscribed angles in a circle central and inscribed angle Tangents to a circle Inscribed angle and tangent to a circle Without using : Law of cosines Define : F the tangent intersection of CirleD and CircleO E intersection of CircleD and OA G intersection […]

Gogeometry Problem 283


From Gogeometry Using : Inscribed angles in a circle central and inscribed angle congruent triangles Similar triangles Tangents to a circle Inscribed angle and tangent to a circle   Without using : Pythagoras Define G the center of CircleG with radius x Define the point E as the only point […]

Gogeometry Problem 276


From Gogeometry Using : Isometric transformation Additional points congruent triangles Define α = ∠BAF and β = ∠DAE α+β+45°=90° => α+β = 45° Define E’ rotation of E anticlockwise with an angle of 90° =>AE=AE’, β = ∠DAE = ∠BAE’ ∠FAE’= α+β = 45° ΔFAE’ is congruent to ΔFAE (SAS) […]

Gogeometry Problem 1076



From Gogeometry   Using : Additional points Inscribed angles in a circle congruent triangles Equilateral triangle CAD intercepts arc CD such as CBD => ∠CAD= ∠CBD Define D’ n AD such as AD’C is congruent to BDC => AD’=f and ∠ACD’= ∠BCD ∠ACD= ∠ACB+ ∠BCD=Π/3+ ∠BCD And ∠ACD= ∠ACD’+ ∠D’CD=∠BCD+ […]

Gogeometry Problem 256


From Gogeomety     Using : Equilateral triangle Congruent triangles Parallelogram ΔACB’ and ΔBCA’ are equilateral => ∠ACB’ =∠BCA’ = Π/3 ∠ACB = ∠ACB’ – ∠B’CB = Π/3 – ∠B’CB ∠B’CA’ = ∠BCA’ – ∠B’CB = Π/3 – ∠B’CB =>∠ACB =∠B’CA’ ΔACB is congruent to ΔB’CA’ (SAS) => BA=B’A’ In […]

Gogeometry Problem 242


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



From Gogeometry Using Pythagoras central and inscribed angle Equilateral triangle DC=DA and ∠ADC= π/3 => ΔADC is equilateral∠CAD= π/3, ∠BAC+∠CAD=∠BAD=75° => ∠BAC=15°∠ACD= π/3, ∠BCA+∠ACD=∠BCD=155° => ∠BCA=75°∠BAC=15° et ∠BCA=75° => ∠ABC= π/2 ΔADC is equilateral and EC=ED => ∠AEC= π/2∠ABC= π/2 et ∠AEC= π/2 => A, B, C and E are […]

Gogeometry Problem 405


From Gogeometry Using Additional points congruent triangles Equilateral triangle Thales theorem ∠ABC = 30°+10°=40°Define F on AC and on the angle bisector of <ABCΔABC isosceles in B => ∠ABF=∠FBC = 40°/2=20°Define I such as ΔAIC is equilateral => ∠IAC=60°ΔABC isosceles in B => ∠ABI=∠CBI = 20°∠ABC = 40° and ΔABC […]

Gogeometry Problem 60


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



From Gogeometry Using Additional points congruent triangles Tangents to a circle Inscribed circle of a triangle Kite GN=GK=rg, DK and DN tangent to CircleG=> DK=DNFL=FM=rf, DL and DM tangent to CircleF=> DL=DMF center of circleF => F is on the angle bisector of ∠ADB=> ∠FDA= ∠FDB=xG center of circleG => […]

Pb 032


From Gogeometry Using Additional points Inscribed angles in a circle central and inscribed angle congruent triangles Concyclic points ∠ABC=∠ADC = π/2 => ABCD are concyclic with center O=> OA=OD=OC=OB Define D’ such as ΔCAD is congruent to ΔCAD’∠CDA=∠CD’A = π/2 =>D’ is on the circle O=> ∠ACD= ∠ACD’= α => […]

Gogeometry Problem 017


From Gogeometry Using Additional points Inscribed angles in a circle congruent triangles Thales theorem Concyclic points ∠DBC = ∠DBA+∠ABC = π/2+∠ABC∠ABG = ∠ABC+∠BCG = ∠ABC+ π/2=> ∠DBC = ∠ABG=> Δ DBC is congruent to Δ ABG (SAS)=> ∠BDC = ∠BAG and DC=AG Define K the intersection of AG and BC=> […]

Gogeometry Problem 497



From Gogeometry Using Additional points Pythagoras congruent triangles Similar triangles Equilateral triangle Let b=AD=DC and d=BCΔACB and ΔBCD are similar since ∠BAC=∠DBC=x and ∠BCD=∠BCAThen AC/BC=CB/CD=AB/BD => 2b/d=d/b => d=b√2 Define C’ such as DC’ ⊥ AC and DC’=b∠BDC=∠ADC – π/4= π – π/4 = 3 π/4∠BDC’=∠BDA + ∠ADC’ = π/4 […]

Gogeometry Problem 001