Properties of Triangles ** * A circle passing through the vertices of the triangle is called circumcircle. * The circle which can be inscribed within the triangle so that it touches all the 3 sides, is called the incircle. * The circle which touches the side BC and also extensions of AB and AC is called escribed circle opposite to angle A. * The circle which touches the side AC and also extensions of AB and BC is called escribed circle opposite to angle B. * The circle which touches the side AB and also extensions of BC and AC is called escribed circle opposite to angle C. * The triangle formed by joining feet of the altitudes drawn from vertices to the opposite sides is called pedal triangle. * Orthocenter/circumcenter of an acute angled triangle lies inside the triangle; that of an obtuse angle triangle lies outside the triangle. & that of right angle triangle lies at the vortex/mid-point of hypotenuse which contain right angle. *Image of the orthocentre (H) with respect to any side of triangle lies on the circumcircle. Here HK=KT where T is the image of H with respect to BC & it lies on circumcircle. * Minimum value of X is 3 i.e when a=b=c; Area= [s(s-a)(s-b)(s-c)]1/2  where s=(a+b+c)/2 or Area=absinC / 2 = bcsinA / 2 =acsinB / 2; sinA = 2*area/bc; sinB=2*area/ac ; sinC=2*area/ab; cosA=(b2 + c2- a2 ) / 2bc ; cosB=(a2 + c2- a2 ) / 2ac ; cosC=(b2 + a2- c2 ) / 2ba sin(A/2) =[ (s-c)(s-b)/bc ]1/2 cos(A/2) =[ s(s-a)/bc ]1/2 Radius of circum-circle= a/2sinA =b/2sinB=c/2sinC = abc/(4*area) Radius of incircle= area / sRadius of escribed circle ra (opposite to angle A)= area/(s-a)Radius of escribed circle rb (opposite to angle B)= area/(s-b) Radius of escribed circle rc(opposite to angle C)= area/(s-c) Circumradius of Pedal triangle= (radius of circumcircle)/2; Length of Median AD= (1/2)*[(2b2 + 2c2- a2]1/2 If H is ortho center,AH=2RcosA, HK=2RcosBcosC; BH=2RcosB;HL=2RcosAcosC,CH=2RcosC;HM=2RcosAcosB; Equn. of Plane passing through A,B,C is of the form Ax+By+Cz=D where A=y2z3-z2y3+y1z2-y2z1+z1y3-z3y1, B=x1z3-x3z1+x2z1-z2x1+x3z2-z3x2, C=x2y3-x3y2+x1y2-y1x2+y1x3-x1y3 & D=x1(y2z3-z2y3)+x2(z1y3-y1z3)+x3(y1z2-z1y2). * BK/CK=(a*a+c*c-b*b) / (a*a+b*b-c*c); *AL / CL =(b*b+c*c-a*a)/(b*b+a*a-c*c); *AM/BM=(c*c+b*b-a*a)/(c*c+a*a-b*b); *acos(B-C)+bcos(C-A)+ccos(A-B)= abc/R2  ; where cos(B-C)=cosBcosC+sinBsinC; * If a rectangle with altitude x is inscribed in triangle ABC with base b and altitude h, the perimeter P of rectangle=2[b(h-x)/x  + x] and area A = b(h-x)*x/h ; for Area A to be extremum, dA/dx=0 which implies b-2bx/h=0 or x=h/2; d2A/dx2 = -2b/h = less than zero which means that extremum is a maxima. Since  dP/dx= 0 yields b=h and does not include x, study the variation of P with variation of x. *Equation of a straight line in 2-D say x-y plane is represented as a linear equation of 2 variables such as Ax+By+C=0 where A,B,C are constants. But in 3 dimensions, the equation is 2 linear equations of 3 variables having constants  say (a3,b3,c3,d3) & (a4,b4,c4,d4). Because one equation represents a plane and a st. line is an intersection between 2 planes. Unsymmetrical Equn. of BC in 3-D (a2x+b2y+c2z+d2=0) &  (a3x+b3y+c3z+d3=0) =>a'x+b'y+c'z+d'=0 where a'=(a2-a3), b'=(b2-b3),c'=(c2-c3) & d'=(d2-d3). Putting the value of co-ordinates -> a'x2+b'y2+c'z2+d' =0 .....(1)  and  a'x3+b'y3+c'z3+d' =0 .....(2) . (1)-(2) =a'dx+b'dy+c'dz=0. Take arbitrary values of b',c'. Then a'=-(b'dy+c'dz)/dx ; If general equation of the plane on which the triangle rests is given by ax+by+cz=d , then a3=a'+a, and a4=a, b3=b'+b and b4=b and c3=c'+c and c4=c; Hence out of the 2 equations of the side of a triangle, 1 equation represents the equation of the plane on which the triangle lies and is common to all the 3 sides. And the second equation represents the unique plane of each side.* throughout the calculations, PI is nowhere used except converting degree to radian since javascript math function of sin, cos are based on radians rather than degree. Suppose, the perimeter of the triangle(p) is converted to an arc with radius x, arc length y and the task is to find out the value of x,y for which area A under the arc shall be maximum. Given p=2x+y or y=p-2x; area A=xy/2=x(p-2x)/2 . Then dA/dx=0 which means x=p/4, y=p/2. d2A/dx2 = -Ve which means above is a case of maximum area. 