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 ABC\triangle ABC we will denote the angles BAC,CBA,ACBBAC, CBA, ACB by A,B,CA, B, C and the corresponsing sides i.e. sides opposite to them by a,b,ca, b, c respectively.

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

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

16.1. Sine Formula or Sine Rule or Law of Sines

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

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

actute angle sine law

From AA draw ADBC.AD \perp BC. From ABD,\triangle ABD,

sinB=ADAB=ADcAD=csinB\sin B = \frac{AD}{AB} = \frac{AD}{c}\Rightarrow AD = c\sin B

From ACD,\triangle ACD,

sinC=ADAC=ADbAD=bsinC\sin C = \frac{AD}{AC} = \frac{AD}{b}\Rightarrow AD = b\sin C

Thus, csinB=bsinCc\sin B = b\sin C

Case II. When C\angle C is obtuse:

obtuse angle sine law

From AA draw ADBC.AD \perp BC. From ABD,\triangle ABD,

sinB=ADAB=ADcAD=csinB\sin B = \frac{AD}{AB} = \frac{AD}{c}\Rightarrow AD = c\sin B

From ACD,\triangle ACD,

sin(πC)=ADAC=ADbAD=bsinC\sin(\pi - C) = \frac{AD}{AC} = \frac{AD}{b}\Rightarrow AD = b\sin C

Thus, csinB=bsinCc\sin B = b\sin C

Case III. When C\angle C is 9090^\circ

right angle sine law

From AA draw ADBC.AD \perp BC. From ABD,\triangle ABD,

sinB=ADAB=ADcAD=csinB\sin B = \frac{AD}{AB} = \frac{AD}{c}\Rightarrow AD = c\sin B

AC=csinB[C\Rightarrow AC = c\sin B[\because C and DD are same points ]]

b=csinBbsin90=csinBb = c\sin B \Rightarrow b\sin90^\circ = c\sin B

bsinC=csinB\Rightarrow b\sin C = c\sin B

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

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

asinA=bsinB\frac{a}{\sin A} = \frac{b}{\sin B}

and thus

ABC,asinA=bsinB=csinC\triangle ABC, \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

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

Case I. When A\angle A is acute.

acute angle sine law

From BDC,\triangle BDC,

sinA=BCBD=a2RasinA=2R\sin A = \frac{BC}{BD} = \frac{a}{2R} \Rightarrow \frac{a}{\sin A} = 2R

Case II. When A\angle A is obtuse.

obtuse angle sine law

From BDC,\triangle BDC,

sin(πA)=BCBD=a2RasinA=2R\sin (\pi - A) = \frac{BC}{BD} = \frac{a}{2R} \Rightarrow \frac{a}{\sin A} = 2R

Case III. When A\angle A is 9090^\circ.

right angle sine law

From BDC,\triangle BDC,

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

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

Similalry, by joining the diameter through AA and OO and through CC and O,O, we can show that bsinB=2R\frac{b}{\sin B} = 2R and csinC=2R\frac{c}{\sin C} = 2R

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

16.2. Tangent Rule

In any ABC,\triangle ABC,

tanBC2=bcb+ccotA2,\tan \frac{B - C}{2} = \frac{b - c}{b + c}\cot \frac{A}{2}, tanAB2=aba+bcotC2\tan \frac{A - B}{2} = \frac{a - b}{a + b}\cot \frac{C}{2} and tanCA2=cac+acotB2\tan \frac{C - A}{2} = \frac{c - a}{c + a}\cot \frac{B}{2}

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

b=KsinB,c=ksinC\therefore b = K\sin B, c = k\sin C

bcb+c=K(sinBsinC)K(sinB+sinC)=2cosB+C2sinBC22sinB+C2cosBC2\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}}

=cotB+C2tanBC2=tanA2tanBC2tanBC2=bcb+ccotA2= \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 ABC,\triangle ABC,

cosA=b2+c2a22bc,cosB=c2+a2b22ca,cosC=a2+b2c22ab\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 C\angle C is acute.

actute angle cosine law

AD=bsinCAD = b\sin C

cosC=CDACCD=bcosCBD=BCCD=abcosC\cos C = \frac{CD}{AC} \Rightarrow CD = b\cos C \Rightarrow BD = BC - CD = a - b\cos C

Case II. When C\angle C is obtuse.

obtuse angle cosine law

AD=bsin(πC)=bsinCAD = b\sin(\pi - C) = b\sin C

cos(πC)=CDACCD=bcosCBD=BC+CD=abcosC\cos (\pi - C) = \frac{CD}{AC} \Rightarrow CD = -b\cos C \Rightarrow BD = BC + CD = a - b\cos C

Case III. When C\angle C is 9090^\circ.

right angle cosine law

Here, CC and DD are same points. AD=AC=b=bsinCAD = AC = b = b\sin C

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

CD=0=bcosCcosC=cos90=0CD = 0 = b\cos C \because \cos C = \cos 90^\circ = 0

BD=BCCD=abcosCBD = BC - CD = a - b\cos C

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

Now, AB2=AD2+BD2AB^2 = AD^2 + BD^2

c2=b2sin2C+(abcosC)2=a2+b22abcosC\Rightarrow c^2 = b^2\sin^2C + (a - b\cos C)^2 = a^2 + b^2 - 2ab\cos C

cosC=a2+b2c22ab\cos C = \frac{a^2 + b^2 - c^2}{2ab}

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

16.4. Projection Formula

In any ABC,\triangle ABC, c=acosB+bcosA,b=ccosA+acosC,a=bcosC+ccosAc = a\cos B + b\cos A, b = c\cos A + a\cos C, a = b\cos C + c\cos A

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

actute angle sine law

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

Case II. When C\angle C is obtuse:

obtuse angle sine law

BC=a=BDCD=ccosBbcos(πC)=ccosB+bcosCBC = a = BD - CD = c\cos B - b\cos(\pi - C) = c\cos B + b\cos C

Case III. When C\angle C is 9090^\circ

right angle sine law

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

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

16.5. Sub Angle Rules

In any ABC,\triangle ABC,

sinA2=(sb)(sc)bc\sin \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}}

cosA2=s(sa)bc\cos \frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}

tanA2=(sb)(sc)s(sa)tan \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}

where 2s=a+b+c2s = a + b + c

Proof:

  1. 2sin2A2=1cosA=1b2+c2a22bc2\sin^2\frac{A}{2} = 1 - \cos A = 1 - \frac{b^2 + c^2 - a^2}{2bc}

    =2bcb2c2+a22bc=a2(bc)22bc=(a+bc)(a+cb)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}

    =(2s2c)(2s2b)2bc=2(sb)(sc)bc= \frac{(2s - 2c)(2s - 2b)}{2bc} = \frac{2(s - b)(s - c)}{bc}

    sin2A2=(sb)(sc)bc\Rightarrow \sin^2\frac{A}{2} = \frac{(s - b)(s - c)}{bc}

    sinA2=±(sb)(sc)bc\Rightarrow \sin \frac{A}{2} = \pm\sqrt{\frac{(s - b)(s - c)}{bc}}

    But A2,\frac{A}{2}, is an acute angle so sinA2>0\sin \frac{A}{2} > 0

    sinA2=(sb)(sc)bc\therefore \sin \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}}

  2. 2cos2A2=1+cosA=1+b2+c2a22bc2\cos^2\frac{A}{2} = 1 + \cos A = 1 + \frac{b^2 + c^2 - a^2}{2bc}

    =(b+c)2a22bc=(a+b+c)(b+ca)2bc= \frac{(b + c)^2 - a^2}{2bc} = \frac{(a + b + c)(b + c - a)}{2bc}

    =2s(2s2a)2bc=2s(sa)bc= \frac{2s(2s - 2a)}{2bc} = \frac{2s(s - a)}{bc}

    cos2A2=s(sa)bc\Rightarrow \cos^2\frac{A}{2} = \frac{s(s - a)}{bc}

    cosA2=±s(sa)bc\cos\frac{A}{2} = \pm\sqrt{\frac{s(s - a)}{bc}}

    But A2\frac{A}{2} is an acute angle so cosA2>0\cos\frac{A}{2} > 0

    cosA2=s(sa)bc\therefore \cos\frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}

  3. From 1 and 2 it follows that tanA2=(sb)(sc)s(sa)\tan\frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}

Similarly, we can prove that sinB2=(sc)(sa)ca,sinC2=(sa)(sb)ab,\sin\frac{B}{2} = \sqrt{\frac{(s - c)(s - a)}{ca}}, \sin\frac{C}{2} = \sqrt{\frac{(s - a)(s - b)}{ab}}, cosB2=s(sb)ca,cosC2=s(sc)ab,\cos\frac{B}{2} = \sqrt{\frac{s(s - b)}{ca}}, \cos\frac{C}{2} = \sqrt{\frac{s(s - c)}{ab}}, tanB2=(sc)(sa)s(sb),tanC2=(sa)(sb)s(sc)\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 ABC\triangle ABC

sinA=2bcs(sa)(sb)(sc)\sin A = \frac{2}{bc}\sqrt{s(s - a)(s - b)(s - c)}

sinB=2cas(sa)(sb)(sc)\sin B = \frac{2}{ca}\sqrt{s(s - a)(s - b)(s - c)}

sinC=2abs(sa)(sb)(sc)\sin C = \frac{2}{ab}\sqrt{s(s - a)(s - b)(s - c)}

Proof: sinA=2sinA2cosA2\sin A = 2\sin\frac{A}{2}\cos\frac{A}{2}

=2(sb)(sc)bcs(sa)bc= 2\sqrt{\frac{(s - b)(s - c)}{bc}}\sqrt{\frac{s(s - a)}{bc}}

=2bcs(sa)(sb)(sc)= \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 ABC,\triangle ABC, then

=12absinC=12bcsinA=12casinB\triangle = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B

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

actute angle sine law

sinC=ADACAD=bsinC\sin C = \frac{AD}{AC} \Rightarrow AD = b\sin C

=12BC×AD=12absinC\triangle = \frac{1}{2}BC\times AD = \frac{1}{2}ab\sin C

Case II. When C\angle C is obtuse:

obtuse angle sine law

sin(πC)=ADACAD=bsinC\sin(\pi - C) = \frac{AD}{AC} \Rightarrow AD = b\sin C

=12BC×AD=12absinC\triangle = \frac{1}{2}BC\times AD = \frac{1}{2}ab\sin C

Case III. When C\angle C is 9090^\circ

right angle sine law

=12BC×AD=12absinC[C=90sinC=1]\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 =12absinC\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 ABC,\triangle ABC, then

=s(sa)(sb)(sc)\triangle = \sqrt{s(s - a)(s - b)(s - c)}

Proof: =12absinC=12ab.2sinC2cosC2\triangle = \frac{1}{2}ab\sin C = \frac{1}{2}ab.2\sin\frac{C}{2}\cos\frac{C}{2}

=ab(sa)(sb)abs(sc)ab=s(sa)(sb)(sc)= 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

=12absinC=12ab.c2R=abc4R\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 ABC,tanA2=(sb)(sc),tanB2=(sa)(sc),tanC2=(sa)(sb),\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}, cotA2=s(sa),cotB2=s(sb),cotC2=s(sc)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: tanA2=(sb)(sc)s(sa)=(sb)2(sc)2s(sa)(sb)(sc)\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)}}

=(sb)(sc)= \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 DD be a point on the side BCBC of a ABC\triangle ABC such that BD:DC=m:nBD:DC = m:n and ADC=θ,BAD=α\angle ADC=\theta, \angle BAD=\alpha and DAC=β,\angle DAC=\beta, then

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

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

ratio of angles

Proof:

  1. ADB=πθ,ABD=π(α+πθ)=θα\angle ADB = \pi - \theta, \angle ABD = \pi - (\alpha + \pi - \theta) = \theta - \alpha

    ACD=π(θ+β)\angle ACD = \pi - (\theta + \beta)

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

    From ADC,DCsin]beta=ADsin[π(θ+β)]\triangle ADC, \frac{DC}{\sin]beta} = \frac{AD}{\sin[\pi - (\theta + \beta)]}

    DCsinβ=ADsin(θ+β)\Rightarrow \frac{DC}{\sin\beta} = \frac{AD}{\sin(\theta + \beta)}

    Dividing, we get

    BDsinβDCsinα=sin(θ+β)sin(θα)\frac{BD\sin\beta}{DC\sin\alpha} = \frac{\sin(\theta + \beta)}{\sin(\theta - \alpha)}

    msinβnsin]α=sin]thetacosβ+cosθcosβsinθcosαcosθsinα\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}

    msinθsinβcosαmcosθsinαsinβ=nsinαsinθcosβ+nsinαcosθsinβ\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 sinαsinβsinθ,\sin\alpha\sin\beta\sin\theta, we get

    mcotαmcotθ=ncotβ+ncotθm\cot\alpha - m\cot\theta = n\cot\beta + n\cot\theta

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

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

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

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

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

    DCsin(θ+C)=ADsinC\Rightarrow \frac{DC}{\sin(\theta + C)} = \frac{AD}{\sin C}

    Dividing, we get

    BDDC.sin(θ+C)sin(θB)=sinCsinB\frac{BD}{DC}.\frac{\sin(\theta + C)}{\sin(\theta - B)} = \frac{\sin C}{\sin B}

    mbsinθcosC+cosθsinCsinθcosBcosθsinB=sinCsinC\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)cotθ=ncotBmcotC(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. RR

Referring to the image of in-circle, IF=r,FAI=A2IF = r, \angle FAI = \frac{A}{2}

From right-angle triangle FIA,sinA2=rAIAI=rcosecA2FIA, \sin\frac{A}{2} = \frac{r}{AI} \Rightarrow AI = r\cosec\frac{A}{2}

Similarly, BI=rcosecB2BI = r\cosec\frac{B}{2} and CI=rcosecC2CI = r\cosec\frac{C}{2}

16.13.1. Orthocenter

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

ortho center

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

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

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

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

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

Similarly, BH=2RcosB,CH=2RcosCBH = 2R\cos B, CH=2R\cos C

16.13.2. Centroid

centroid

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

From geometry, AB2+AC2=2DB2+2AD2AB^2 + AC^2 = 2DB^2 + 2AD^2

c2+b22.(a2)2+2AD2\Rightarrow c^2 + b^2 2.\left(\frac{a}{2}\right)^2 + 2AD^2

2AD2=2b2+2c2a22\Rightarrow 2AD^2 = \frac{2b^2 + 2c^2 - a^2}{2}

AG:GD=2:1\because AG:GD = 2:1 [property of centroid that it divides median in the ratio of 2:12:1 ]

AG=23AD=132b2+2c2a2AG = \frac{2}{3}AD = \frac{1}{3}\sqrt{2b^2 + 2c^2 - a^2}

Similarly, BG=132a2+2c2b2,CG=132a2+2b2c2BG = \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 BAD=β\angle BAD=\beta and CAD=γ,\angle CAD=\gamma, then we have

sinγsinC=DCAD\frac{\sin \gamma}{\sin C} = \frac{DC}{AD}

sinγ=DC.sinCAD=asinC2b2+2c2a2\sin \gamma = \frac{DC.\sin C}{AD} = \frac{a\sin C}{\sqrt{2b^2 + 2c^2 - a^2}}

Similarly, sinβ=asinB2b2+2c2a2\sin\beta = \frac{a\sin B}{\sqrt{2b^2 + 2c^2 - a^2}}

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

sinθ=2bsinC2b2+2c2a2\sin\theta = \frac{2b\sin C}{2b^2 + 2c^2 - a^2}

16.14. Esccribed Triangles

escribed circles

Let II be the incenter and I1,I2I_1,I_2 and I3I_3 be centers of excircles opposite to vertices A,BA, B and CC respectively. We know that ICIC will bisect the ACB\angle ACB and I1CI_1C will bisect the external angles at BB and CC produced by extending the sides i.e. BCM\angle BCM as shown in figure.

ICI1=ICB+ICB\therefore ICI_1 = \angle ICB + \angle I_CB

=12ACB+12BCM=90= \frac{1}{2}ACB + \frac{1}{2}\angle BCM = 90^\circ

Similarly, ICI2\angle ICI_2 and ICI3\angle ICI_3 will be right angles.

Hence, I1CI2I_1C_I2 is perpendicular to IC.IC. Similarly, I2AI3I_2AI_3 is perpendicular to IA,IA, and I3BI1I_3BI_1 is perpendicular to IB.IB.

We also see that IAIA and I1AI_1A both bisect A\angle A so I1IAI_1IA is a straight line. Similarly, I2IBI_2IB and I3ICI_3IC are straight lines.

The I1I2I3\triangle I_1I_2I_3 is called excentric triangle of ABC\triangle ABC.

16.15. Distance between Orthocenter and Circumcenter

Let OO be circumcenter, OFABOF\perp AB and HH be orthocenter. Then,

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

Let BLBL perpendicular to ACAC so it will pass through H.H.

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

OAH=AOAFHAL=A(1802C)=CB\angle OAH = A - \angle OAF - \angle HAL = A - (180^\circ - 2C) = C - B

Also, OA=ROA = R and HA=2RcosAHA = 2R\cos A

OH2=OA2+HA22OA.HA.cosOAH=R2+4R2cos2A4R2cosAcos(CB)OH^2 = OA^2 + HA^2 - 2OA.HA.\cos OAH = R^2 + 4R^2\cos^2A - 4R^2\cos A\cos(C - B)

=R2+4R2cosA[cosAcos(CB)]=R28R2cosAcosBcosC= R^2 + 4R^2\cos A[\cos A - \cos(C - B)] = R^2 - 8R^2\cos A\cos B\cos C

OH=R18cosAcosBcosCOH = R\sqrt{1 - 8\cos A\cos B\cos C}

16.16. Distance between Incenter and Circumcenter

Let OO be orthocenter and OFAB.OF\perp AB. Let II be the incenter and ICAB.IC\perp AB.

OAF=90COAI=IAFOAF=A290+C=CB2\angle OAF = 90^\circ - C \therefore \angle OAI = \angle IAF - OAF = \frac{A}{2} - 90^\circ + C = \frac{C - B}{2}

Also, AI=IEsinA2=rsinA2=4RsinB2sinC2AI = \frac{IE}{\sin\frac{A}{2}} = \frac{r}{\sin\frac{A}{2}} = 4R\sin\frac{B}{2}\sin\frac{C}{2}

OI2=OA2+AI22.OA.AI.cosOAI\therefore OI^2 = OA^2 + AI^2 - 2.OA.AI.\cos OAI

=R2+16R2sin2B2sin2C28R2sinB2sinC2cosCB2= 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=R18sinA2sinB2sinC2OI = R\sqrt{1 - 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}

=R22Rr= \sqrt{R^2 - 2Rr}

16.17. Area of a Cyclid Quadrilateral

cyclic quadrilateral

If a,b,c,da, b, c, d be the sides and ss the subperimeter of a cyclic quadrilateral, then its area is (sa)(sb)(sc)(sd)\sqrt{(s - a)(s - b)(s - c)(s - d)}

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

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

Applying cosine law in ABC,cosB=a2+b2AC22abAC2=a2+b22abcosB\triangle ABC, \cos B = \frac{a^2 + b^2 - AC^2}{2ab} \Rightarrow AC^2 = a^2 + b^2 - 2ab\cos B

Similarly, in ACD,AC2=c2+d22cdcos(πB)=c2+d2+2cdcosB\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 cosB=a2+b2c2d22(ab+cd)\cos B = \frac{a^2 + b^2 - c^2 - d^2}{2(ab + cd)}

Area of quadrilateral ABCDABCD = Area of ABC\triangle ABC + Area of ACD\triangle ACD

=12adsinB+12cdsin(πB)=12(ab+cd)sinB= \frac{1}{2}ad\sin B + \frac{1}{2}cd \sin(\pi - B) = \frac{1}{2}(ab + cd)\sin B

Also, sin2B=1cos2B=1[a2+b2c2d22(ab+cd)]2\sin^2B = 1 - \cos^2B = 1 - \left[\frac{a^2 + b^2 - c^2 - d^2}{2(ab + cd)}\right]^2

=(2ab+2cd+a2+b2c2d2)(2ab+2cda2b2+c2+d2)4(ab+cd)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}

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

=4(sa)(sb)(sc)(sd)4(ab+cd)2= \frac{4(s - a)(s - b)(s - c)(s - d)}{4(ab + cd)^2}

Thus, area of quadrilateral =(sa)(sb)(sc)(sd)= \sqrt{(s - a)(s - b)(s - c)(s - d)}