The Trigonometry of the Triangle

Draw the diameter $BX$ through $B$

Angle $BAX = 90$ degrees and $angle AXB = angle ACB$

From the diagram it can be seen that $c = 2R \sin C$

$\therefore\;\;\;\;\;\frac{c}{\sin\,C} = 2R$

Therefore by symmetry:

$\frac{a}{\sin\,A} = \frac{b}{\sin\,B} = \frac{c}{\sin\,C} = 2R$

If $A$ is obtuse, angle $BXC = 180 - A$

$a = 2R\,\sin\,(180^0 - A) = 2R\,\sin\,A$

$\frac{a}{\sin\,A} = \frac{b}{\sin\,B} = \frac{c}{\sin\,C} = 2R$

Then $CD = b\;\cos\,C$

therefore $BD = a - b\;\cos\,C$

Using Pythagoras:

$h^2 = b^2 - (b\;\cos\,C)^2$

and $\;\;\;\;\;\;h^2 = c^2 - (a - b\;\cos\,C)^2$

therefore $b^2 - b^2\,\cos^2\,C = c^2 - a^2 - b^2\;\cos^2\,C + 2ab\;\cos\,C$

or $\;\;\;\;\;\;\;c^2 = a^2 + b^2 - 2ab\;\cos\,C$

$CD = b\;\cos\,(180^0 - C)$

$BD = a + b\;\cos\,(180^0 - C)$

$\therefore\;\;\;\;\;h^2 = b^2 - b^2\;\cos^2\,(180^0 - C)$

and $\;\;\;\;\;h^2 = c^2 - \left(a + b\;\cos\,[180^0 - C] \right)^2$

$b^2 - b^2\;\cos^2(180^0 - C)\;= c^2 - a^2 - b^2\;\cos^2\,(180^0 - C)$ $- 2ab\;\cos\,(180^0 - C)$

or $\;\;\;\;\;c^2 = a^2 + b^2 + 2ab\;\cos\,(180^0 - C)$

Thus $\;\;\;\;\;c^2 = a^2 + b^2 - 2ab\;\cos\,C$

Note It should be noted that the same equation can be applied in both cases.

Using the sine formula

$\frac{a - b}{a + b} = \frac{2R\;\sin\,A - 2R\;\sin\,B}{2R\;\sin\,A + 2R\;\sin\,B}= \frac{2\;\cos\,\frac{1}{2}(A + B)\times \sin\;\frac{1}{2}(A - B)}{2\;\sin\,\frac{1}{2}(A + B)\times \cos\,\frac{1}{2}\,(A - B)}$

$= \frac{\tan\,\frac{1}{2}(A - B)}{\tan\,\frac{1}{2}(A + B)} = \frac{\tan\,\frac{1}{2}(A - B)}{\cot\,\frac{1}{2}C}$

as $C$ and $(A + B)$ are complements

$\therefore\;\;\;\;\;\frac{a - b}{a + b}\times \cot\,\frac{C}{2} = \tan\,\frac{A - B}{2}$

$c = AN + NB$

$\therefore\;\;\;\;\;c = a\;\cos\,B + b\;\cos\,A$

$\therefore\;\;\;\;\;\;\cos\;\frac{C}{2} = \sqrt{\frac{s(s - c)}{ab}}$

$\therefore\;\;\;\;\;\;\sin\,\frac{C}{2} = \sqrt{\frac{(s - a)(s - b)}{ab}}$

By division $\;\;\;\;\;\;\tan\,\frac{C}{2} = \sqrt{\frac{(s - a)(s - b)}{s(s - c)}}$

But

Since the area

The area of the triangle can be written as :

$S = \frac{1}{2}\;ab\times 2\;\sin\,\frac{1}{2}C\times \cos\,\frac{1}{2}\,C=$

$=ab\;\sqrt{\frac{(s - a)(s - b)}{ab}}\sqrt{\frac{s(s - c)}{ab}}$

$AB^2 + AC^2 = 2AA'^2 + 2BA'^2$

$\therefore\;\;\;\;\;\;b^2 + c^2 = 2AA'^2 + \frac{1}{2}\;a^2$

$\therefore\;\;\;\;\;4AA'^2 = 2b^2 + 2c^2\;-a^2$

$\therefore\;\;\;\;\;\;AA' = \frac{1}{2}\sqrt{2b^2 + 2c^2\;-a^2}$

Also $\;\;\;\;\;\;AG = \frac{2}{3}AA' = \frac{\sqrt{2b^2 + 2c^2\;-a^2}}{3}$

$\frac{AH}{\sin\;ABH} = \frac{c}{\sin\;AHB}$
But $\displaystyle \angle\;ABH = (90^0 - A)$ and $\displaystyle \angle\;AHB = (180^0 - C)$
$\therefore\;\;\;\;\;\frac{AH}{\cos\,A} = \frac{c}{\sin\,C} = 2R$

$\therefore\;\;\;\;\;\;AH = 2R\;\cos\,A$

Similarly $\;\;\;\;\;BH = 2R\;\cos\,B$

But $\;\;\;\;\;\angle\;HBD = 90^0 - C$ Therefore $HD = 2R\;\cos\,B\;\cos\,C$

Let $\triangle ABC$ and $X$ such as $AX$ is the bisector of the angle $\angle BAC$

As $AX$ bisects the angle $BAC$ internally

But $\displaystyle \sin\,A = \sin\,\frac{1}{2}\,A\;\cos\,\frac{1}{2}\,A$

$\frac{BX}{XC} = \frac{c}{b}$

And since $\displaystyle BX + XC = a$

$BX = \frac{c\,a}{b + c}$

Applying the sine formula to $AXB$

$\frac{AX}{\sin\,B} = \frac{BX}{\sin\,\frac{1}{2}A}= \frac{a\;c}{(b + c)\;\sin\,\frac{1}{2}A}$

$AX = \frac{a\;c\;\sin\,B}{(b + c)\;\sin\,\frac{1}{2}A}= \frac{c\;b\;\sin\,A}{(b + c)\;\sin\,\frac{1}{2}A}$ Therefore
The Angle bisector $\displaystyle AX = \frac{2\;cb\,\cos\,\frac{1}{2}A}{b + c}$

Since $FHDB$ is cyclic

$H\hat{D}F = H\hat{B}F = 90^0 - A$

And Since $DHEC$ is cyclic$F\hat{D}E = H\hat{C}E = 90^0 - A$

By addition

$F\hat{D}E = 180^0 - 2A$

$H$ is the incentre of the Pedal Triangle and the angles are given by:

$(180^0 - 2A);(180^0 - 2B);(180^0 - 2C)$

Note The sides of the Pedal Triangle are $a \cos A$ ; $b \cos B$ ; and $c \cos C$

Since $DBEA$ is cyclic, $\angle AED = \angle ABD = B$

Since $BEFC$ is cyclic, $\angle FEA = \angle FBC = B$

And therefore $\angle DEF = 2B$

Similarly $\angle EFD = 2C$

By subtraction $E\hat{D}F = 180^0 - 2B - 2C = 180^0 - 2(180^0 - A) = 2A - 180^0$

$A$ is the Incentre of the Triangle $\triangle HBC$ and the angles of the Pedal Triangle are:

$(2A - 180^0)\,;\,(2B)\,;(2C)$

Using the sine formula for triangle

$AFE$:$\frac{EF}{\sin\,EAF} = \frac{AF}{\sin\,AEF}$

Hence

$\frac{EF}{\sin\,EAF} = \frac{b\,\cos\;(180 - A)}{\sin\,HBC}$ Thus $EF\;= - 2R\;\cos\,A\;\sin\,A = - a\;\cos\,A$ or $- R\;\sin\,2A$

The sides of the Pedal Triangle are $a \cos A$ ; $b \cos B$ ; $c \cos C$

Note It is worth knowing that in the case of either an acute or an obtuse angle triangle, the four points $A,B,C$ and $H$ are the three ex-centre and incentre of the Pedal Triangle.

The Area of the triangle $\triangle ABC$ is the sum of the areas of the triangles $AIB$; $AIC$; $BIC$.

$S_{AIB} = \frac{1}{2}\times AB\times Radius = \frac{1}{2}\;c\;r$

Similar equations can be written for triangles $AIC$ and $BIC$

Therefore the area of triangle $ABC$ is given by:

$S_{ABC}= r\times \frac{1}{2}(a + b + c) = r\times s$ where $s$ is the semi perimeter
If $X , Y , Z$ are the points of contact between the triangle and circle, then

$BX = BZ$ ; $CX = CY$ ; $AY = AZ$ and the semi circumference of the triangle( $s$ ) is given by:

$BX + XC + AY = s = AY + BC$ but $BC = a$

$\therefore\;\;\;\;\;\;AY = s - a$

A similar relationship exists for $AZ$; $BZ$ etc.

For the triangle $AZI$

$r = AI \;\sin\frac{1}{2}A$

Applying the sine formula to triangle $AIB$

$\frac{AI}{\sin\;\frac{1}{2}B} = \frac{c}{\sin\,AIB} = \frac{c}{\sin\,\left(180^0 - \frac{A + B}{2} \right)}$

$= \frac{c}{\sin\,\frac{1}{2}\,(A + B)} = \frac{c}{\cos\,\frac{1}{2}\,C}$

$= \frac{2R\;\sin\,C}{\cos\,\frac{1}{2}\,C} = 4R\;\sin\,\frac{1}{2}\,C$

$\therefore\;\;\;\;\;AI = 4R\;\sin\,\frac{1}{2}\,B\times \sin\,\frac{1}{2}\,C$ Thus $r = 4R\;\sin\,\frac{1}{2}\,A\,\times \sin\,\frac{1}{2}\, B\times \sin\,\frac{1}{2}\,C$

Let $P$, $Q$, $R$, be the points of contact between the lines which make up the sides of the triangle $ABC$ and the circle.

Equating the areas of triangles $ABC$, $AIB$, $AIC$, $BIC$. we get:

$\Delta ABC = \Delta AIB + \Delta AIC - \Delta BIC$

This can be re-written in terms of $r$ the radius of the ex-circle

$\Delta ABC = \frac{1}{2}\,c\,r + \frac{1}{2}\,b\,r - \frac{1}{2}\;a\,r = \frac{1}{2}\,r\,(b + c - a)$

$2\,s = a + b + c$

$\therefore\;\;\;\;\;\;\Delta ABC = r(s - a)$

Thus the radius of the ex-circle, $r$ is given by the equation:

$r = \frac{\Delta ABC}{(s - a)}$

By inspection $AR = AQ$ ; $BR = BP$ ; $CQ = CP$.

$AB + BP = AC + CP = \frac{1}{2}\;perimeter = s$

$BP = s - c\;\;\;and\;\;\;PC = s - b$

From triangle $AIR$

$r = AI\;\sin\,\frac{1}{2}\,A$

Using the sine formula in triangle $ABI$

$\frac{AI}{\sin\,ABI} = \frac{c}{\sin\,AIB}$

$\frac{AI}{\sin\;(90^0 + \frac{1}{2}B)} = \frac{c}{\sin\;(90^0 - \frac{1}{2}B - \frac{1}{2}A)}$

$\frac{AI}{\cos\,\frac{1}{2}B} = \frac{c}{\sin\frac{1}{2}C} = \frac{2R\;\sin\,C}{\sin\frac{1}{2}C}$

$= 4R\;\cos\,\frac{1}{2}C$

$AI = 4R\,\cos\,\frac{1}{2}\,B\;\cos\,\frac{1}{2}\,C$

$r = 4R\;\sin\,\frac{1}{2}\;A\times \cos\,\frac{1}{2}\,B\times \cos\,\frac{1}{2}\,C$

Since $AI_2\,,\,AI_3$ are the external bisectors of the angle $A$ , the line $I_2 A$ is a straight line as is also the line $A\,I\,I_1$( the Internal bisector)
As the external and internal bisectors of an angle are perpendicular, $AI_1$ is perpendicular to $I_2\,I_3$
Therefore the triangle $ABC$ is the Pedal Triangle of $I_1,I_2,I_3$ and $I$ is the orthocentre.
Since $I_1\,BC = 90^0 - \frac{1}{2}\,B$ and $I_1CB = 90^0 - \frac{1}{2}C$

$BI_1C = \frac{1}{2}(B + C) = 90^0 - \frac{1}{2}\,A$ therefore $BC = I_2I_3\;\cos(90^0 - \frac{1}{2}A)$ Thus $I_2I_3 = \frac{a}{\sin\frac{1}{2}A} = 2R\;\frac{\sin\;A}{\sin\frac{1}{2}A}$

Therefore the Length of the side of the Triangle through $A$ is :

$I_2I_3 = 4R\;\cos\;\frac{1}{2}A$

The nine point circle of $I_1I_2I_3$ will pass through the feet of its altitudes $ABC$. The radius of the its nine point circle is therefor $R$. But since the radius of a nine point circle is half that of the circum-circle, the radius of the circum-cirle of $I_1I_2I_3$ is $2R$

$\therefore\;\;\;\;\;\;II_1\; = 2(2R)\;\cos\,(90^0 - \frac{1}{2}A)$ Hence $II_1 = 4R\;\sin \,\frac{1}{2}A$