# Euclide Geometry Problem From Gogeometry Using : Additional points Pythagoras Solution of a previous problem Tangents to a circle Proposition 13 Euclide Book II R radius of CircleO, and r of CircleF Define ro radius of CircleO , rd of CircleD, rc of CircleC and rf of CircleF Define G intersection of OA […]

## Gogeometry Problem 285 From Gogeometry Using : Pythagoras   Define h=OF OA=OB=R D, E and C are aligned ΔCOD is right in O => DC^2=OC^2+OD^2 OD=r+h => (1) : (r+R/2)^2=(R/2)^2+(r+h)^2 => (1) : (r+R)^2=R^2+4(r+h)^2 OB=OA => R=2r+h => r+h=R-r (1) : (r+R)^2=R^2+4(R-r)^2 => r^2+2rR+R^2=R^2+4(R^2-2rR+r^2) => 6rR= 3r^2 Therefore 3R=r

## Gogeometry Problem 284 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 : Inscribed angles in a circle congruent triangles Concyclic points ABCD is a rhombus => ΔACD is congruent to ΔACB => ∠ACD = ∠ACB => ΔGCD is congruent to ΔGCB (SAS) => ∠GDC = ∠GBC ABCD is a rhombus =>CE=CD => ΔDCE is isosceles in C => […]

## Gogeometry Problem 354 From Gogeometry Using : Pythagoras Area of a triangle Define c=AB S=[ABC]=a.b/2=c.h/2 => a.b=c.h c=a.b/h => c^2=a^2.b^2/h^2 Pythagoras => a^2+b^2=c^2 => a^2+b^2=a^2.b^2/h^2 dividing by a^2.b^2 => 1/b^2 + 1/a^2 = 1/h^2

## Gogeometry Problem 268 From Gogeometry Using : Pythagoras Area of a triangle S=[ABC] S=a.b/2 But also S=c.h/2 Therefore a.b=c.h

## Gogeometry Problem 267 From Gogeometry Using : Pythagoras (1) ΔABC : AC^2+BC^2=c^2 (2) ΔAHC : AC^2 =h^2+n^2 (3) ΔBHC : BC^2=h^2+m^2 c=m+n, c^2=(m+n)^2=m^2+n^2+2mn (1,2,3) h^2+n^2+h^2+m^2=m^2+n^2+2mn 2h^2=2mn Therefore h^2=m.n

## Gogeometry Problem 266 From Gogeometry Using : Additional points congruent triangles Solution of a previous problem 218 DB=GF+FJ From problem 217, ΔJBF is congruent to ΔHDE (AAA) => JF=EH Therefore a+b=h

## Gogeometry Problem 218 From Gogeometry Using : Isometric transformation Additional points congruent triangles Define J in FG such as JG ⊥ JB => JB=GD ED//BF => ∠JBF=∠HDE In ΔHDE, ∠HDE + ∠DEH =90° => in ΔJBF, ∠JBF + ∠BFJ =90° =>∠JFB = ∠HED FB=DE => ΔJBF is congruent to ΔHDE (AAA) => JB=HD […]