Mathematics Chapter 7 Triangles

In ΔABC and ΔABD,

AC = AD (Given)

∠CAB = ∠DAB (AB bisects ∠A)

AB = AB (Common)

∴ ΔABC ≅ ΔABD (By SAS congruence rule)

∴ BC = BD (By CPCT)

Therefore, BC and BD are of equal lengths.

In ΔABD and ΔBAC,

AD = BC (Given)

∠DAB = ∠CBA (Given)

AB = BA (Common)

∴ ΔABD ≅ ΔBAC (By SAS congruence rule)

∴ BD = AC (By CPCT)

And, ∠ABD = ∠BAC (By CPCT)

In ΔBOC and ΔAOD,

∠BOC = ∠AOD (Vertically opposite angles)

∠CBO = ∠DAO (Each 90º)

BC = AD (Given)

∴ ΔBOC ≅ ΔAOD (AAS congruence rule)

∴ BO = AO (By CPCT)

⇒ CD bisects AB.

In ΔABC and ΔCDA,

∠BAC = ∠DCA (Alternate interior angles, as p || q)

AC = CA (Common)

∠BCA = ∠DAC (Alternate interior angles, as l || m)

∴ ΔABC ≅ ΔCDA (By ASA congruence rule)

In ΔAPB and ΔAQB,

∠APB = ∠AQB (Each 90º)

∠PAB = ∠QAB (l is the angle bisector of ∠A)

AB = AB (Common)

∴ ΔAPB ≅ ΔAQB (By AAS congruence rule)

∴ BP = BQ (By CPCT)

Or, it can be said that B is equidistant from the arms of ∠A.

It is given that ∠BAD = ∠EAC

∠BAD + ∠DAC = ∠EAC + ∠DAC

∠BAC = ∠DAE

In ΔBAC and ΔDAE,

AB = AD (Given)

∠BAC = ∠DAE (Proved above)

AC = AE (Given)

∴ ΔBAC ≅ ΔDAE (By SAS congruence rule)

∴ BC = DE (By CPCT)

It is given that ∠EPA = ∠DPB

⇒ ∠EPA + ∠DPE = ∠DPB + ∠DPE

⇒ ∠DPA = ∠EPB

In ΔDAP and ΔEBP,

∠DAP = ∠EBP (Given)

AP = BP (P is mid-point of AB)

∠DPA = ∠EPB (From above)

∴ ΔDAP ≅ ΔEBP (ASA congruence rule)

∴ AD = BE (By CPCT)

(i) It is given that in triangle ABC, AB = AC

⇒ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are equal)

⇒ $\frac{1}{2}$∠ACB = $\frac{1}{2}$∠ABC

⇒ ∠OCB = ∠OBC

⇒ OB = OC (Sides opposite to equal angles of a triangle are also equal)

(ii) In ΔOAB and ΔOAC,

AO =AO (Common)

AB = AC (Given)

OB = OC (Proved above)

Therefore, ΔOAB ≅ ΔOAC (By SSS congruence rule)

⇒ ∠BAO = ∠CAO (CPCT)

⇒ AO bisects ∠A.

In ΔADC and ΔADB,

AD = AD (Common)

∠ADC =∠ADB (Each 90º)

CD = BD (AD is the perpendicular bisector of BC)

∴ ΔADC ≅ ΔADB (By SAS congruence rule)

∴AB = AC (By CPCT)

Therefore, ABC is an isosceles triangle in which AB = AC

In ΔAEB and ΔAFC,

∠AEB and ∠AFC (Each 90º)

∠A = ∠A (Common angle)

AB = AC (Given)

∴ ΔAEB ≅ ΔAFC (By AAS congruence rule)

⇒ BE = CF (By CPCT)

(i) In ΔABE and ΔACF,

∠AEB = ∠AFC (Each 90º)

∠A = ∠A (Common angle)

BE = CF (Given)

∴ ΔABE ≅ ΔACF (By AAS congruence rule)

(ii) It has already been proved that

ΔABE ≅ ΔACF

⇒ AB = AC (By CPCT)

Let us join AD.

In ΔABD and ΔACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common side)

∴ ΔABD $\cong$ΔACD (By SSS congruence rule)

⇒ ∠ABD = ∠ACD (By CPCT)

In ΔABC,

AB = AC (Given)

⇒ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are also equal)

In ΔACD,

AC = AD

⇒ ∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are also equal)

In ΔBCD,

∠ABC + ∠BCD + ∠ADC = 180º (Angle sum property of a triangle)

⇒ ∠ACB + ∠ACB +∠ACD + ∠ACD = 180º

⇒ 2(∠ACB + ∠ACD) = 180º

⇒ 2(∠BCD) = 180º

⇒ ∠BCD = 90º

It is given that

AB = AC

⇒ ∠C = ∠B (Angles opposite to equal sides are also equal)

In ΔABC,

∠A + ∠B + ∠C = 180º (Angle sum property of a triangle)

⇒ 90º + ∠B + ∠C = 180º

⇒ 90º + ∠B + ∠B = 180º

⇒ 2 ∠B = 90º

⇒ ∠B = 45º

∴ ∠B = ∠C = 45º

Let us consider that ABC is an equilateral triangle.

Therefore, AB = BC = AC

AB = AC

⇒ ∠C = ∠B (Angles opposite to equal sides of a triangle are equal)

Also,

AC = BC

⇒ ∠B = ∠A (Angles opposite to equal sides of a triangle are equal)

Therefore, we obtain

∠A = ∠B = ∠C

In ΔABC,

∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠A + ∠A = 180°

⇒ 3∠A = 180°

⇒ ∠A = 60°

⇒ ∠A = ∠B = ∠C = 60°

Hence, in an equilateral triangle, all interior angles are of measure 60º.

(i) In ΔABD and ΔACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common)

∴ ΔABD ≅ ΔACD (By SSS congruence rule)

⇒ ∠BAD = ∠CAD (By CPCT)

⇒ ∠BAP = ∠CAP …. (1)

(ii) In ΔABP and ΔACP,

AB = AC (Given)

∠BAP = ∠CAP [From equation (1)]

AP = AP (Common)

∴ ΔABP ≅ ΔACP (By SAS congruence rule)

⇒ BP = CP (By CPCT) … (2)

(iii) From equation (1),

∠BAP = ∠CAP

Hence, AP bisects ∠A.

In ΔBDP and ΔCDP,

BD = CD (Given)

DP = DP (Common)

BP = CP [From equation (2)]

∴ ΔBDP ≅ ΔCDP (By S.S.S. Congruence rule)

⇒ ∠BDP = ∠CDP (By CPCT) … (3)

Hence, AP bisects ∠D.

(iv) ΔBDP ≅ ΔCDP

∴ ∠BPD = ∠CPD (By CPCT) …. (4)

∠BPD + ∠CPD = 180$°$ (Linear pair angles)

∠BPD + ∠BPD = 180$°$

2∠BPD = 180$°$ [From equation (4)]

∠BPD = 90$°$ … (5)

From equations (2) and (5), it can be said that AP is the perpendicular bisector of BC.

(i) In ΔBAD and ΔCAD,

∠ADB = ∠ADC (Each 90º as AD is an altitude)

AB = AC (Given)

AD = AD (Common)

∴ΔBAD ≅ ΔCAD (By RHS Congruence rule)

⇒ BD = CD (By CPCT)

Hence, AD bisects BC.

(ii) Also, by CPCT,

∠BAD = ∠CAD

Hence, AD bisects ∠A.

(i) In ΔABC, AM is the median to BC.

∴ BM = $\frac{1}{2}$BC

In ΔPQR, PN is the median to QR.

∴ QN = $\frac{1}{2}$QR

However, BC = QR

∴ $\frac{1}{2}$BC = $\frac{1}{2}$QR

⇒ BM = QN … (1)

In ΔABM and ΔPQN,

AB = PQ (Given)

BM = QN [From equation (1)]

AM = PN (Given)

∴ ΔABM ≅ ΔPQN (SSS congruence rule)

∠ABM = ∠PQN (By CPCT)

∠ABC = ∠PQR … (2)

(ii) In ΔABC and ΔPQR,

AB = PQ (Given)

∠ABC = ∠PQR [From equation (2)]

BC = QR (Given)

⇒ ΔABC ≅ ΔPQR (By SAS congruence rule)

In ΔBEC and ΔCFB,

∠BEC = ∠CFB (Each 90°)

BC = CB (Common)

BE = CF (Given)

∴ ΔBEC ≅ ΔCFB (By RHS congruency)

⇒ ∠BCE = ∠CBF (By CPCT)

∴ AB = AC (Sides opposite to equal angles of a triangle are equal)

Hence, ΔABC is isosceles.

In ΔAPB and ΔAPC,

∠APB = ∠APC (Each 90º)

AB =AC (Given)

AP = AP (Common)

∴ ΔAPB ≅ ΔAPC (Using RHS congruence rule)

⇒ ∠B = ∠C (By using CPCT)

Let us consider a right-angled triangle ABC, right-angled at B.

In ΔABC,

∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)

∠A + 90º + ∠C = 180°

∠A + ∠C = 90°

Hence, the other two angles have to be acute (i.e., less than 90º).

∴ ∠B is the largest angle in ΔABC.

⇒ ∠B > ∠A and ∠B > ∠C

⇒ AC > BC and AC > AB

[In any triangle, the side opposite to the larger (greater) angle is longer.]

Therefore, AC is the largest side in ΔABC.

However, AC is the hypotenuse of ΔABC. Therefore, hypotenuse is the longest side in a right-angled triangle.

In the given figure,

∠ABC + ∠PBC = 180° (Linear pair)

⇒ ∠ABC = 180° − ∠PBC … (1)

Also,

∠ACB + ∠QCB = 180°

∠ACB = 180° − ∠QCB … (2)

As ∠PBC < ∠QCB,

⇒ 180º − ∠PBC > 180º − ∠QCB

⇒ ∠ABC > ∠ACB [From equations (1) and (2)]

⇒ AC > AB (Side opposite to the larger angle is larger.)

In ΔAOB,

∠B < ∠A

⇒ AO < BO (Side opposite to smaller angle is smaller) … (1)

In ΔCOD,

∠C < ∠D

⇒ OD < OC (Side opposite to smaller angle is smaller) … (2)

On adding equations (1) and (2), we obtain

AO + OD < BO + OC

AD < BC

Let us join AC.

In ΔABC,

AB < BC (AB is the smallest side of quadrilateral ABCD)

∴ ∠2 < ∠1 (Angle opposite to the smaller side is smaller) … (1)

In ΔADC,

AD < CD (CD is the largest side of quadrilateral ABCD)

∴ ∠4 < ∠3 (Angle opposite to the smaller side is smaller) … (2)

On adding equations (1) and (2), we obtain

∠2 + ∠4 < ∠1 + ∠3

⇒ ∠C < ∠A

⇒ ∠A > ∠C

Let us join BD.

In ΔABD,

AB < AD (AB is the smallest side of quadrilateral ABCD)

∴ ∠8 < ∠5 (Angle opposite to the smaller side is smaller) … (3)

In ΔBDC,

BC < CD (CD is the largest side of quadrilateral ABCD)

∴ ∠7 < ∠6 (Angle opposite to the smaller side is smaller) … (4)

On adding equations (3) and (4), we obtain

∠8 + ∠7 < ∠5 + ∠6

⇒ ∠D < ∠B

⇒ ∠B > ∠D

As PR > PQ,

∴ ∠PQR > ∠PRQ (Angle opposite to larger side is larger) … (1)

PS is the bisector of ∠QPR.

∴∠QPS = ∠RPS … (2)

∠PSR is the exterior angle of ΔPQS.

∴ ∠PSR = ∠PQR + ∠QPS … (3)

∠PSQ is the exterior angle of ΔPRS.

∴ ∠PSQ = ∠PRQ + ∠RPS … (4)

Adding equations (1) and (2), we obtain

∠PQR + ∠QPS > ∠PRQ + ∠RPS

⇒ ∠PSR > ∠PSQ [Using the values of equations (3) and (4)]

Let us take a line l and from point P (i.e., not on line l), draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.

In ΔPNM,

∠N = 90º

∠P + ∠N + ∠M = 180º (Angle sum property of a triangle)

∠P + ∠M = 90º

Clearly, ∠M is an acute angle.

∴ ∠M < ∠N

⇒ PN < PM (Side opposite to the smaller angle is smaller)

Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to them.

Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together.

In ΔABC, we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA of this triangle. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of ΔABC.

The point which is equidistant from all the sides of a triangle is called the incentre of the triangle. Incentre of a triangle is the intersection point of the angle bisectors of the interior angles of that triangle.

Here, in ΔABC, we can find the incentre of this triangle by drawing the angle bisectors of the interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. Therefore, I is the point equidistant from all the sides of ΔABC.

Maximum number of persons can approach the ice-cream parlour if it is equidistant from A, B and C. Now, A, B and C form a triangle. In a triangle, the circumcentre is the only point that is equidistant from its vertices. So, the ice-cream parlour should be set up at the circumcentre O of ΔABC.

In this situation, maximum number of persons can approach it. We can find circumcentre O of this triangle by drawing perpendicular bisectors of the sides of this triangle.

It can be observed that hexagonal-shaped rangoli has 6 equilateral triangles in it.

Area of ΔOAB $=\frac{\sqrt{3}}{4}{\left(side\mathit{\right)}}^{\mathit{ }\mathit{2}}\mathit{=}\frac{\sqrt{3}}{4}{\mathit{\left(}\mathit{5}\mathit{\right)}}^{\mathit{ }\mathit{2}}$

$=\frac{\sqrt{3}}{4}\left(25\right)=\frac{25\sqrt{3}}{4}c{m}^2$

Area of hexagonal-shaped rangoli$=6\times \frac{25\sqrt{3}}{4}$$=\frac{75\sqrt{3}}{2}c{m}^{\mathit{2}}$

Area of equilateral triangle having its side as 1 cm $\frac{\sqrt{3}}{4}{\left(1\right)}^2=\frac{\sqrt{3}}{4}c{m}^2$

Star-shaped rangoli has 12 equilateral triangles of side 5 cm in it.

Area of star-shaped rangoli = $12\times \frac{\sqrt{3}}{4}\times {\left(5\right)}^2$ $=75\sqrt{3}$

Therefore, star-shaped rangoli has more equilateral triangles in it.