# 16. Properties of Triangles¶

In this chapter we will study the relations between the sides and trigonometrical ratios of the angles of a triangle. We already know that a triangle has three sides and three angles. In a $\triangle ABC$ we will denote the angles $BAC, CBA, ACB$ by $A, B, C$ and the corresponsing sides i.e. sides opposite to them by $a, b, c$ respectively.

Thus, $BC = a, AC = b, AB = c$

We will also denote the radius of the circumcircle of the $\triangle ABC$ by $R$ and the area by $\triangle.$ We also know some basic properties of a triangle for example, $A + B + C = 180^\circ$ and $a + b > c, b + c > a, c + a > b$

## 16.1. Sine Formula or Sine Rule or Law of Sines¶

In $\triangle ABC, \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Proof: Case I. When $\angle C$ is acute:

From $A$ draw $AD \perp BC.$ From $\triangle ABD,$

$\sin B = \frac{AD}{AB} = \frac{AD}{c}\Rightarrow AD = c\sin B$

From $\triangle ACD,$

$\sin C = \frac{AD}{AC} = \frac{AD}{b}\Rightarrow AD = b\sin C$

Thus, $c\sin B = b\sin C$

Case II. When $\angle C$ is obtuse:

From $A$ draw $AD \perp BC.$ From $\triangle ABD,$

$\sin B = \frac{AD}{AB} = \frac{AD}{c}\Rightarrow AD = c\sin B$

From $\triangle ACD,$

$\sin(\pi - C) = \frac{AD}{AC} = \frac{AD}{b}\Rightarrow AD = b\sin C$

Thus, $c\sin B = b\sin C$

Case III. When $\angle C$ is $90^\circ$

From $A$ draw $AD \perp BC.$ From $\triangle ABD,$

$\sin B = \frac{AD}{AB} = \frac{AD}{c}\Rightarrow AD = c\sin B$

$\Rightarrow AC = c\sin B[\because C$ and $D$ are same points $]$

$b = c\sin B \Rightarrow b\sin90^\circ = c\sin B$

$\Rightarrow b\sin C = c\sin B$

Thus, from all cases we have established that $\frac{b}{\sin B} = \frac{c}{\sin C}$

Similarly by drawing perpendicular from $C$ to $AB,$ we can prove that

$\frac{a}{\sin A} = \frac{b}{\sin B}$

and thus

$\triangle ABC, \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

In $\triangle ABC, \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$ where $R$ is the radius of the circumcircle of $\triangle ABC$

Case I. When $\angle A$ is acute.

From $\triangle BDC,$

$\sin A = \frac{BC}{BD} = \frac{a}{2R} \Rightarrow \frac{a}{\sin A} = 2R$

Case II. When $\angle A$ is obtuse.

From $\triangle BDC,$

$\sin (\pi - A) = \frac{BC}{BD} = \frac{a}{2R} \Rightarrow \frac{a}{\sin A} = 2R$

Case III. When $\angle A$ is $90^\circ$.

From $\triangle BDC,$

$a = BC = 2R \Rightarrow \frac{a}{\sin A} = 2R$

Thus, in all three cases $\frac{a}{\sin A} = 2R$

Similalry, by joining the diameter through $A$ and $O$ and through $C$ and $O,$ we can show that $\frac{b}{\sin B} = 2R$ and $\frac{c}{\sin C} = 2R$

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

## 16.2. Tangent Rule¶

In any $\triangle ABC,$

$\tan \frac{B - C}{2} = \frac{b - c}{b + c}\cot \frac{A}{2},$ $\tan \frac{A - B}{2} = \frac{a - b}{a + b}\cot \frac{C}{2}$ and $\tan \frac{C - A}{2} = \frac{c - a}{c + a}\cot \frac{B}{2}$

Proof: By sine formula, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = K$ (say)

$\therefore b = K\sin B, c = k\sin C$

$\Rightarrow \frac{b - c}{b + c} = \frac{K(\sin B - \sin C)}{K(\sin B + \sin C)} = \frac{2\cos\frac{B + C}{2}\sin\frac{B - C}{2}}{2\sin \frac{B + C}{2}\cos\frac{B - C}{2}}$

$= \cot\frac{B + C}{2}\tan\frac{B - C}{2} = \tan\frac{A}{2}\tan \frac{B - C}{2} \Rightarrow \tan \frac{B - C}{2} = \frac{b - c}{b + c}\cot \frac{A}{2}$

Similarly, we can prove remaining two equations.

## 16.3. Cosine Formula or Cosine Rule¶

In any $\triangle ABC,$

$\cos A = \frac{b^2 + c^2 - a^2}{2bc}, \cos B = \frac{c^2 + a^2 - b^2}{2ca}, \cos C = \frac{a^2 + b^2 - c^2}{2ab}$

Case I. When $\angle C$ is acute.

$AD = b\sin C$

$\cos C = \frac{CD}{AC} \Rightarrow CD = b\cos C \Rightarrow BD = BC - CD = a - b\cos C$

Case II. When $\angle C$ is obtuse.

$AD = b\sin(\pi - C) = b\sin C$

$\cos (\pi - C) = \frac{CD}{AC} \Rightarrow CD = -b\cos C \Rightarrow BD = BC + CD = a - b\cos C$

Case III. When $\angle C$ is $90^\circ$.

Here, $C$ and $D$ are same points. $AD = AC = b = b\sin C$

$\angle C = 90^\circ\therefore \sin C = 1$

$CD = 0 = b\cos C \because \cos C = \cos 90^\circ = 0$

$BD = BC - CD = a - b\cos C$

Thus, in all cases $AD = b\sin C$ and $BD = a - b\cos C$

Now, $AB^2 = AD^2 + BD^2$

$\Rightarrow c^2 = b^2\sin^2C + (a - b\cos C)^2 = a^2 + b^2 - 2ab\cos C$

$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

Similarly, we can prove it for $\angle A$ and $\angle B.$

## 16.4. Projection Formula¶

In any $\triangle ABC,$ $c = a\cos B + b\cos A, b = c\cos A + a\cos C, a = b\cos C + c\cos A$

Proof: Case I. When $\angle C$ is acute:

$BC = a = BD + CD = c\cos B + b\cos C$

Case II. When $\angle C$ is obtuse:

$BC = a = BD - CD = c\cos B - b\cos(\pi - C) = c\cos B + b\cos C$

Case III. When $\angle C$ is $90^\circ$

$BD = a = BC + CD = c\cos B + b\cos C[\because C=90^\circ \cos C = 0]$

Thus, in all cases $a = b\cos C + c\cos B,$ similarly we can prove for other sides as well.

## 16.5. Sub Angle Rules¶

In any $\triangle ABC,$

$\sin \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}}$

$\cos \frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}$

$tan \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}$

where $2s = a + b + c$

Proof:

1. $2\sin^2\frac{A}{2} = 1 - \cos A = 1 - \frac{b^2 + c^2 - a^2}{2bc}$

$= \frac{2bc - b^2 - c^2 + a^2}{2bc} = \frac{a^2 - (b - c)^2}{2bc} = \frac{(a + b - c)(a + c - b)}{2bc}$

$= \frac{(2s - 2c)(2s - 2b)}{2bc} = \frac{2(s - b)(s - c)}{bc}$

$\Rightarrow \sin^2\frac{A}{2} = \frac{(s - b)(s - c)}{bc}$

$\Rightarrow \sin \frac{A}{2} = \pm\sqrt{\frac{(s - b)(s - c)}{bc}}$

But $\frac{A}{2},$ is an acute angle so $\sin \frac{A}{2} > 0$

$\therefore \sin \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}}$

2. $2\cos^2\frac{A}{2} = 1 + \cos A = 1 + \frac{b^2 + c^2 - a^2}{2bc}$

$= \frac{(b + c)^2 - a^2}{2bc} = \frac{(a + b + c)(b + c - a)}{2bc}$

$= \frac{2s(2s - 2a)}{2bc} = \frac{2s(s - a)}{bc}$

$\Rightarrow \cos^2\frac{A}{2} = \frac{s(s - a)}{bc}$

$\cos\frac{A}{2} = \pm\sqrt{\frac{s(s - a)}{bc}}$

But $\frac{A}{2}$ is an acute angle so $\cos\frac{A}{2} > 0$

$\therefore \cos\frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}$

3. From 1 and 2 it follows that $\tan\frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}$

Similarly, we can prove that $\sin\frac{B}{2} = \sqrt{\frac{(s - c)(s - a)}{ca}}, \sin\frac{C}{2} = \sqrt{\frac{(s - a)(s - b)}{ab}},$ $\cos\frac{B}{2} = \sqrt{\frac{s(s - b)}{ca}}, \cos\frac{C}{2} = \sqrt{\frac{s(s - c)}{ab}},$ $\tan\frac{B}{2} = \sqrt{\frac{(s - c)(s - a)}{s(s - b)}}, \tan\frac{C}{2} = \sqrt{\frac{(s - a)(s - b)}{s(s - c)}}$

## 16.6. Sines of Angles in Terms of Sides¶

In any $\triangle ABC$

$\sin A = \frac{2}{bc}\sqrt{s(s - a)(s - b)(s - c)}$

$\sin B = \frac{2}{ca}\sqrt{s(s - a)(s - b)(s - c)}$

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

Proof: $\sin A = 2\sin\frac{A}{2}\cos\frac{A}{2}$

$= 2\sqrt{\frac{(s - b)(s - c)}{bc}}\sqrt{\frac{s(s - a)}{bc}}$

$= \frac{2}{bc}\sqrt{s(s - a)(s - b)(s - c)}$

Similarly, we can prove it for other angles.

## 16.7. Area of a Triangle¶

If $\triangle$ denotes the area of $\triangle ABC,$ then

$\triangle = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B$

Proof: Case I. When $\angle C$ is acute:

$\sin C = \frac{AD}{AC} \Rightarrow AD = b\sin C$

$\triangle = \frac{1}{2}BC\times AD = \frac{1}{2}ab\sin C$

Case II. When $\angle C$ is obtuse:

$\sin(\pi - C) = \frac{AD}{AC} \Rightarrow AD = b\sin C$

$\triangle = \frac{1}{2}BC\times AD = \frac{1}{2}ab\sin C$

Case III. When $\angle C$ is $90^\circ$

$\triangle = \frac{1}{2}BC\times AD = \frac{1}{2}ab\sin C[\because C=90^\circ \therefore \sin C = 1]$

Thus, in all cases $\triangle = \frac{1}{2}ab\sin C$

Similarly, we can prove two other formulas.

### 16.7.1. Area in Terms of Sides¶

If $\triangle$ be the area of any $\triangle ABC,$ then

$\triangle = \sqrt{s(s - a)(s - b)(s - c)}$

Proof: $\triangle = \frac{1}{2}ab\sin C = \frac{1}{2}ab.2\sin\frac{C}{2}\cos\frac{C}{2}$

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

#### 16.7.1.1. Area in Terms of Radius of Circumcircle¶

$\triangle = \frac{1}{2}ab\sin C = \frac{1}{2}ab.\frac{c}{2R} = \frac{abc}{4R}$

## 16.8. Tangent and Cotangent of Sub-angles of a Triangle¶

In any $\triangle ABC, \tan\frac{A}{2} = \frac{(s - b)(s - c)}{\triangle}, \tan \frac{B}{2}= \frac{(s - a)(s - c)}{\triangle}, \tan \frac{C}{2} = \frac{(s - a)(s - b)}{\triangle},$ $\cot\frac{A}{2} = \frac{s(s - a)}{\triangle}, \cot\frac{B}{2} = \frac{s(s - b)}{\triangle}, \cot\frac{C}{2} = \frac{s(s - c)}{triangle}$

Proof: $\tan\frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} = \sqrt{\frac{(s - b)^2(s - c)^2}{s(s - a)(s - b)(s - c)}}$

$= \frac{(s - b)(s - c)}{\triangle}$

Similarly, we can prove it for other angles and cotangents.

## 16.9. Dividing a Side in a Ratio¶

If $D$ be a point on the side $BC$ of a $\triangle ABC$ such that $BD:DC = m:n$ and $\angle ADC=\theta, \angle BAD=\alpha$ and $\angle DAC=\beta,$ then

1. $(m + n)\cot\theta = m\cot\alpha - n\cot\beta$

2. $(m + n)\cot\theta = n\cot B - m\cot C$

Proof:

1. $\angle ADB = \pi - \theta, \angle ABD = \pi - (\alpha + \pi - \theta) = \theta - \alpha$

$\angle ACD = \pi - (\theta + \beta)$

From $\triangle ABC, \frac{BD}{\sin\alpha} = \frac{AD}{\sin(\theta - \alpha)}$

From $\triangle ADC, \frac{DC}{\sin]beta} = \frac{AD}{\sin[\pi - (\theta + \beta)]}$

$\Rightarrow \frac{DC}{\sin\beta} = \frac{AD}{\sin(\theta + \beta)}$

Dividing, we get

$\frac{BD\sin\beta}{DC\sin\alpha} = \frac{\sin(\theta + \beta)}{\sin(\theta - \alpha)}$

$\Rightarrow \frac{m\sin\beta}{n\sin]\alpha} = \frac{\sin]theta\cos\beta + \cos\theta\cos\beta}{\sin\theta\cos\alpha - \cos\theta\sin\alpha}$

$\Rightarrow m\sin\theta\sin\beta\cos\alpha - m\cos\theta\sin\alpha\sin\beta = n\sin\alpha\sin\theta\cos\beta + n\sin\alpha\cos\theta\sin\beta$

Dividing both sides by $\sin\alpha\sin\beta\sin\theta,$ we get

$m\cot\alpha - m\cot\theta = n\cot\beta + n\cot\theta$

$\Rightarrow (m + n)\cot\theta = m\cot\alpha - n\cot\beta$

2. $\angle BAD = 180^\circ - (180^\circ - \theta + B) = \theta - B$

$\angle DAC = 180^\circ - (\theta + C)$

From $\triangle BAD, \frac{BD}{\sin(\theta - B)} = \frac{AD}{\sin B}$

From $\triangle ADC, \frac{DC}{\sin(180^\circ - (\theta + C))} = \frac{AD}{\sin C}$

$\Rightarrow \frac{DC}{\sin(\theta + C)} = \frac{AD}{\sin C}$

Dividing, we get

$\frac{BD}{DC}.\frac{\sin(\theta + C)}{\sin(\theta - B)} = \frac{\sin C}{\sin B}$

$\Rightarrow \frac{m}{b}\frac{\sin\theta\cos C + \cos\theta\sin C}{\sin\theta\cos B - \cos\theta\sin B} = \frac{\sin C}{\sin C}$

Proceeding like previous proof, we get

$(m + n)\cot\theta = n\cot B - m\cot C$

## 16.13. Distannces of Circum-center, In-center, Orthocenter and Centroid from Vertices¶

We have already shown that for circum-center distance is equal to circum-radius i.e. $R$

Referring to the image of in-circle, $IF = r, \angle FAI = \frac{A}{2}$

From right-angle triangle $FIA, \sin\frac{A}{2} = \frac{r}{AI} \Rightarrow AI = r\cosec\frac{A}{2}$

Similarly, $BI = r\cosec\frac{B}{2}$ and $CI = r\cosec\frac{C}{2}$

### 16.13.1. Orthocenter¶

Orthocenter is point of intersection of perpendiculars from a vertex to opposite side.

Let the orthocenter be $H$ which is intersection of perpendiculars from any vertex to opposite side.

From right angle triangle $AEB, \cos A= \frac{AE}{AB} \Rightarrow AE = c\cos A$

From right angle triangle $ADC, \angle DAC = 90^\circ - C$

From right angle triangle $AEH, \cos(90^\circ - C) = \frac{AE}{AH}$

$\Rightarrow AH = \frac{c\cos A}{\sin C} = 2R\cos A$

Similarly, $BH = 2R\cos B, CH=2R\cos C$

### 16.13.2. Centroid¶

Let $G$ be the centroid. Since, it is the point of intersection of medians, it will lie on median $AD.$

From geometry, $AB^2 + AC^2 = 2DB^2 + 2AD^2$

$\Rightarrow c^2 + b^2 2.\left(\frac{a}{2}\right)^2 + 2AD^2$

$\Rightarrow 2AD^2 = \frac{2b^2 + 2c^2 - a^2}{2}$

$\because AG:GD = 2:1$ [property of centroid that it divides median in the ratio of $2:1$ ]

$AG = \frac{2}{3}AD = \frac{1}{3}\sqrt{2b^2 + 2c^2 - a^2}$

Similarly, $BG = \frac{1}{3}\sqrt{2a^2 + 2c^2 - b^2}, CG = \frac{1}{3}\sqrt{2a^2 + 2b^2 - c^2}$

#### 16.13.2.1. Angle Made by Medians with Sides¶

If $\angle BAD=\beta$ and $\angle CAD=\gamma,$ then we have

$\frac{\sin \gamma}{\sin C} = \frac{DC}{AD}$

$\sin \gamma = \frac{DC.\sin C}{AD} = \frac{a\sin C}{\sqrt{2b^2 + 2c^2 - a^2}}$

Similarly, $\sin\beta = \frac{a\sin B}{\sqrt{2b^2 + 2c^2 - a^2}}$

If $\angle ADC$ be $\theta,$ then we have

$\sin\theta = \frac{2b\sin C}{2b^2 + 2c^2 - a^2}$

## 16.14. Esccribed Triangles¶

Let $I$ be the incenter and $I_1,I_2$ and $I_3$ be centers of excircles opposite to vertices $A, B$ and $C$ respectively. We know that $IC$ will bisect the $\angle ACB$ and $I_1C$ will bisect the external angles at $B$ and $C$ produced by extending the sides i.e. $\angle BCM$ as shown in figure.

$\therefore ICI_1 = \angle ICB + \angle I_CB$

$= \frac{1}{2}ACB + \frac{1}{2}\angle BCM = 90^\circ$

Similarly, $\angle ICI_2$ and $\angle ICI_3$ will be right angles.

Hence, $I_1C_I2$ is perpendicular to $IC.$ Similarly, $I_2AI_3$ is perpendicular to $IA,$ and $I_3BI_1$ is perpendicular to $IB.$

We also see that $IA$ and $I_1A$ both bisect $\angle A$ so $I_1IA$ is a straight line. Similarly, $I_2IB$ and $I_3IC$ are straight lines.

The $\triangle I_1I_2I_3$ is called excentric triangle of $\triangle ABC$.

## 16.15. Distance between Orthocenter and Circumcenter¶

Let $O$ be circumcenter, $OF\perp AB$ and $H$ be orthocenter. Then,

$\angle OAF = 90^\circ - \angle AOF = 90^\circ - C$

Let $BL$ perpendicular to $AC$ so it will pass through $H.$

$\angle HAL = 90^\circ - C$

$\angle OAH = A - \angle OAF - \angle HAL = A - (180^\circ - 2C) = C - B$

Also, $OA = R$ and $HA = 2R\cos A$

$OH^2 = OA^2 + HA^2 - 2OA.HA.\cos OAH = R^2 + 4R^2\cos^2A - 4R^2\cos A\cos(C - B)$

$= R^2 + 4R^2\cos A[\cos A - \cos(C - B)] = R^2 - 8R^2\cos A\cos B\cos C$

$OH = R\sqrt{1 - 8\cos A\cos B\cos C}$

## 16.16. Distance between Incenter and Circumcenter¶

Let $O$ be orthocenter and $OF\perp AB.$ Let $I$ be the incenter and $IC\perp AB.$

$\angle OAF = 90^\circ - C \therefore \angle OAI = \angle IAF - OAF = \frac{A}{2} - 90^\circ + C = \frac{C - B}{2}$

Also, $AI = \frac{IE}{\sin\frac{A}{2}} = \frac{r}{\sin\frac{A}{2}} = 4R\sin\frac{B}{2}\sin\frac{C}{2}$

$\therefore OI^2 = OA^2 + AI^2 - 2.OA.AI.\cos OAI$

$= R^2 + 16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} - 8R^2\sin\frac{B}{2}\sin\frac{C}{2}\cos\frac{C - B}{2}$

$OI = R\sqrt{1 - 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}$

$= \sqrt{R^2 - 2Rr}$

## 16.17. Area of a Cyclid Quadrilateral¶

If $a, b, c, d$ be the sides and $s$ the subperimeter of a cyclic quadrilateral, then its area is $\sqrt{(s - a)(s - b)(s - c)(s - d)}$

Proof: Let $ABCD$ be the cyclic quadrilateral having sides $AB = a, BC = b, CD = c$ and $AD = d$

Since opposing angles of a quadrilateral are complementary, therefore $B + D = A + C = \pi$

Applying cosine law in $\triangle ABC, \cos B = \frac{a^2 + b^2 - AC^2}{2ab} \Rightarrow AC^2 = a^2 + b^2 - 2ab\cos B$

Similarly, in $\triangle ACD, AC^2 = c^2 + d^2 -2cd\cos(\pi - B) = c^2 + d^2 + 2cd\cos B$

From these two equations, we get $\cos B = \frac{a^2 + b^2 - c^2 - d^2}{2(ab + cd)}$

Area of quadrilateral $ABCD$ = Area of $\triangle ABC$ + Area of $\triangle ACD$

$= \frac{1}{2}ad\sin B + \frac{1}{2}cd \sin(\pi - B) = \frac{1}{2}(ab + cd)\sin B$

Also, $\sin^2B = 1 - \cos^2B = 1 - \left[\frac{a^2 + b^2 - c^2 - d^2}{2(ab + cd)}\right]^2$

$= \frac{(2ab + 2cd + a^2 + b^2 - c^2 - d^2)(2ab + 2cd - a^2 - b^2 + c^2 + d^2)}{4(ab + cd)^2}$

$= \frac{[(a + b)^2 -(c - d)^2][(c + d)^2 - (a - b)^2]}{4(ab + cd)^2}$

$= \frac{4(s - a)(s - b)(s - c)(s - d)}{4(ab + cd)^2}$

Thus, area of quadrilateral $= \sqrt{(s - a)(s - b)(s - c)(s - d)}$