# Triangles

## Class 9 NCERT Maths

### NCERT

1   In quadrilateral $ACBD, AC = AD$ and $AB$ bisects $\angle A$ (See the given figure). Show that $\Delta ABC \cong \Delta ABD$. What can you say about $BC$ and $BD$?

##### Solution :

In $\Delta ABC$ and $\Delta CDA,$$\\ \angle BAC = \angle DCA (Alternate interior angles, as p \parallel q)\\ AC = CA (Common)\\ \angle BCA = \angle DAC (Alternate interior angles, as l \parallel m)\\ \therefore \Delta ABC \cong \Delta CDA (By ASA congruence rule) 5 Line l is the bisector of an angle \angle A and \angle B is any point on l. BP and BQ are perpendiculars from B to the arms of \angle A (see the given figure). Show that:\\ (i) \Delta APB \cong \Delta AQB$$\\$ (ii)$BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.

##### Solution :

It is given that $\angle EPA = \angle DPB$$\\ \angle EPA + \angle DPE = \angle DPB + \angle DPE \\ \therefore \angle DPA = \angle EPB \\ In \Delta DAP and \Delta EBP,$$\\$ $\angle DAP = \angle EBP$ (Given)$\\$ $AP = BP$ ($P$ is mid-point of $AB$)$\\$ $\angle DPA = \angle EPB$ (From above)$\\$ $\therefore \Delta DAP \cong \Delta EBP (ASA$ congruence rule)$\\$ $\therefore AD = BE$ (By $CPCT$)

8   In right triangle $ABC$, right angled at $C, M$ is the mid-point of hypotenuse $AB. C$ is joined to $M$ and produced to a point $D$ such that $DM = CM$. Point $D$ is joined to point $B$ (see the given figure). Show that:$\\$ (i) $\Delta AMC \cong \Delta BMD$$\\ (ii) \angle DBC is a right angle.\\ (iii) \Delta DBC \cong \Delta ACB$$\\$ (iv)$CM = \dfrac{1}{2}AB$

(i) In $\Delta AMC$ and $\Delta BMD,$ $\\$ $AM = BM (M$ is the mid-point of $AB)$ $\\$ $\angle AMC = \angle BMD$ (Vertically opposite angles)$\\$ $CM = DM$ (Given)$\\$ $\therefore AMC \cong \Delta BMD$ (By $SAS$ congruence rule)$\\$ $\therefore AC = BD$ (By $CPCT)$ $\\$ And, $\angle ACM = \angle BDM$ (By $CPCT$)$\\$ (ii) $\angle ACM = \angle BDM$ $\\$ However, $\angle ACM$ and $\angle BDM$ are alternate interior angles.$\\$ Since alternate angles are equal,$\\$ It can be said that $DB || AC$$\\ \angle DBC + \angle ACB = 180^o (Co-interior angles)\\ \angle DBC + 90^o = 180^o \\ \therefore \angle DBC = 90^o \\ (iii) In \Delta DBC and \Delta ACB, \\ DB = AC (Already proved)\\ \angle DBC = \angle ACB (Each 90 )\\ BC = CB (Common)\\ \therefore \Delta DBC \cong \Delta ACB (SAS congruence rule)\\ (iv) \Delta DBC \cong \Delta ACB \\ AB = DC (By CPCT)\\ AB = 2 CM \\ \therefore CM=\dfrac{1}{2}AB 9 In an isosceles triangle ABC, with AB = AC, the bisectors of \angle B and \angle C intersect each other at O. Join A to O. Show that:\\ (i) OB = OC (ii) AO bisects \angle A ##### Solution : (i) It is given that in triangle ABC, AB = AC \\ \angle ACB = \angle ABC (Angles opposite to equal sides of a triangle are equal)\\ \dfrac{1}{2}\angle ACB =\dfrac{1}{2} \angle ABC \\ \angle OCB = \angle OBC \\ \therefore OB = OC (Sides opposite to equal angles of a triangle are also equal)\\ (ii) In \Delta OAB and Delta OAC, \\ AO =AO (Common)\\ AB = AC (Given)\\ OB = OC (Proved above)\\ Therefore, \Delta OAB \cong \Delta OAC (By SSS congruence rule)\\ \angle BAO = \angle CAO (CPCT) \\ \therefore AO bisects \angle A. 10 In \Delta ABC, AD is the perpendicular bisector of BC (see the given figure). Show that \Delta ABC is an isosceles triangle in which AB = AC. ##### Solution : In \Delta ADC and \Delta ADB,$$\\$ $AD = AD$ (Common)$\\$ $\angle ADC =\angle ADB$(Each $90^o)$ $\\$ $CD = BD (AD$ is the perpendicular bisector of $BC)$ $\\$ $\therefore \Delta ADC \cong \Delta ADB$(By $SAS$ congruence rule)$\\$ $\therefore AB = AC$ (By $CPCT$) $\\$ Therefore, $ABC$ is an isosceles triangle in which $AB = AC.$

11   $ABC$ is an isosceles triangle in which altitudes $BE$ and $CF$ are drawn to equal sides $AC$ and $AB$ respectively (see the given figure). Show that these altitudes are equal.

##### Solution :

In $\Delta AEB$ and $\delta AFC,$ $\\$ $\angle AEB$ and $\angle AFC$ (Each $90^o$)$\\$ $\angle A = \angle A$(Common angle)$\\$ $AB = AC$ (Given)$\\$ $\therefore \Delta AEB \cong \Delta AFC$(By $AAS$ congruence rule) $\\$ $\therefore BE = CF$ (By $CPCT$)

12   $ABC$ is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal (see the given figure). Show that$\\$ (i) $\Delta ABE \cong \Delta ACF$ $\\$ (ii) $AB = AC,$ i.e., $ABC$ is an isosceles triangle.

##### Solution :

(i) In $\Delta ABE$ and $\Delta ACF,$ $\\$ $\angle ABE$ and $\angle ACF$ (Each $90^o$)$\\$ $\angle A = \angle A$ (Common angle)$\\$ $BE = CF$ (Given)$\\$ $\therefore ABE \cong \Delta ACF$ (By $AAS$ congruence rule)$\\$ (ii) It has already been proved that $\Delta ABE \cong \Delta ACF$ $\\$ $\therefore AB = AC$ (By $CPCT$)

13   $ABC$ and $DBC$ are two isosceles triangles on the same base $BC$ (see the given figure). Show that $\angle ABD = \angle ACD.$

##### Solution :

Let us join $AD.$ $\\$ In $\Delta ABD$ and $\Delta ACD,$ $\\$ $AB = AC$ (Given)$\\$ $BD = CD$ (Given)$\\$ $AD = AD$ (Common side)$\\$ $\therefore \Delta ABD \cong \Delta ACD$ (By $SSS$ congruence rule) $\\$ $\therefore \angle ABD = \angle ACD$ (By $CPCT$)

14   $\Delta ABC$ is an isosceles triangle in which $AB = AC$ . Side $BA$ is produced to $D$ such that $AD = AB$ (see the given figure). Show that $\angle BCD$ is a right angle.

##### Solution :

In $\delta ABC,$ $\\$ $AB = AC$ (Given)$\\$ $\therefore \angle ACB = \angle ABC$ (Angles opposite to equal sides of a triangle are also equal)$\\$ In $\Delta ACD,$ $\\$ $AC = AD$ $\\$ $\therefore \angle ADC = \angle ACD$ (Angles opposite to equal sides of a triangle are also equal)$\\$ In $\Delta BCD,$ $\\$ $\angle ABC + \angle BCD + \angle ADC = 180^o$ (Angle sum property of a triangle)$\\$ $\angle ACB + \angle ACB +\angle ACD + \angle ACD = 180^o$ $\\$ $2(\angle ACB + \angle ACD) = 180^o$ $\\$ $2(\angle BCD) = 180^o$ $\\$ $\therefore \angle BCD = 90^o$

15   $ABC$ is a right angled triangle in which $\angle A = 90^o$ and $AB = AC$ . Find $\angle B$ and $\angle C.$

##### Solution :

It is given that $\\$ $AB = AC$ $\\$ $\therefore \angle C = \angle B$ (Angles opposite to equal sides are also equal)$\\$ In $\Delta ABC,$ $\\$ $\angle A + \angle B + \angle C = 180^o$ (Angle sum property of a triangle)$\\$ $90^o + \angle B + \angle C = 180^o$ $\\$ $90^o + \angle B + \angle B = 180^o$ $\\$ $2\angle B = 90^o$ $\\$ $\angle B = 45^o$ $\\$ $\therefore \angle B = \angle C = 45^o$ $\\$

16   Show that the angles of an equilateral triangle are $60^o$ each.

##### Solution :

Let us consider that $ABC$ is an equilateral triangle.$\\$ Therefore,$AB = BC = AC$ $\\$ $AB = AC$ $\\$ $\therefore \angle C = \angle B$ (Angles opposite to equal sides of a triangle are equal)$\\$ Also,$\\$ $AC = BC$ $\\$ $\therefore \angle B = \angle A$ (Angles opposite to equal sides of a triangle are equal)$\\$ Therefore, we obtain $\angle A = \angle B = \angle C$ $\\$ In $\Delta ABC,$ $\\$ $\angle A + \angle B + \angle C = 180^o$ $\\$ $\angle A + \angle A + \angle A = 180^o$ $\\$ $3\angle A = 180^o$ $\\$ $\angle A = 60^o$ $\\$ $\therefore \angle A = \angle B = \angle C = 60^o$ $\\$ Hence, in an equilateral triangle, all interior angles are of measure $60^o$.

17   $\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see the given figure). If $AD$ is extended to intersect $BC$ at $P,$ show that $\\$ (i) $\delta ABD \cong \Delta ACD$ $\\$ (ii) $\Delta ABP \cong \Delta ACP$ $\\$ (iii) $AP$ bisects $\angle A$ as well as $\angle D$ .$\\$ (iv) $AP$ is the perpendicular bisector of $BC.$

##### Solution :

(i) In $\Delta ABD$ and $\Delta ACD,$ $\\$ $AB = AC$ (Given)$\\$ $BD = CD$ (Given)$\\$ $AD = AD$ (Common)$\\$ $\Delta ABD \cong \Delta ACD$ (By $SSS$ congruence rule) $\\$ $\angle BAD = \angle CAD$ (By $CPCT$)$\\$ $\angle BAP = \angle CAP .... (1)$ $\\$ (ii) In $\Delta ABP$ and $\Delta ACP,$ $\\$ $AB = AC$ (Given)$\\$ $\angle BAP = \angle CAP$ [From equation (1)]$\\$ $AP = AP$ (Common)$\\$ $\therefore ABP \cong \Delta ACP$ (By $SAS$ congruence rule) $\\$ $\therefore BP = CP$ (By $CPCT$) ... (2)$\\$ (iii) From Equation (1), $\angle BAP = \angle CAP$ $\\$ Hence, $AP$ bisects $\angle A.$ $\\$ In $\Delta BDP$ and $\Delta CDP,$ $\\$ $BD = CD$ (Given)$\\$ $DP = DP$ (Common)$\\$ $BP = CP$ [From equation (2)]$\\$ $\therefore \Delta BDP \cong \Delta CDP$ (By $SSS$ Congruence rule) $\\$ $\therefore \angle BDP = \angle CDP$ (By $CPCT$) ... (3)$\\$ Hence, $AP$ bisects $\angle D.$ $\\$ (iv) $\Delta BDP \cong \Delta CDP$ $\\$ $\therefore \angle BPD = \angle CPD$ (By $CPCT$) .... (4)$\\$ $\angle BPD + \angle CPD = 180^o$ (Linear pair angles)$\\$ $\angle BPD + \angle BPD = 180^o$ $\\$ $2\angle BPD = 180^o$ [From Equation (4)]$\\$ $\angle BPD = 90^o ... (5)$ $\\$ From Equations (2) and (5), it can be said that $AP$ is the perpendicular bisector of $BC.$

18   $AD$ is an altitude of an isosceles triangles $ABC$ in which $AB = AC$. Show that $\\$ (i) $AD$ bisects $BC$$\\ (ii) AD bisects \angle A. ##### Solution : (i) In \Delta BAD and \Delta CAD, \\ \angle ADB = \angle ADC (Each 90^o as AD is an altitude)\\ AB = AC (Given)\\ AD = AD (Common)\\ \therefore \Delta BAD \cong \Delta CAD (By RHS Congruence rule) \\ \therefore BD = CD (By CPCT)\\ Hence, AD bisects BC. \\ (ii) Also, by CPCT, \angle BAD = \angle CAD \\ Hence, AD bisects \angle A 19 Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \Delta PQR (see the given figure). Show that:\\ (i) \Delta ABM \cong \Delta PQN \\ (ii) \Delta ABC \cong \Delta PQR ##### Solution : (i) In \Delta ABC, AM is the median to BC.\\ \therefore BM = \dfrac{1}{2} BC \\ In \Delta PQR, PN is the median to QR. \\ \therefore QN = \dfrac{1}{2 }QR \\ However, BC = QR \\ \therefore \dfrac{1}{2} BC = \dfrac{1}{2} QR \\ \therefore BM=QN ...(1) \\ In \Delta ABM and \Delta PQN, \\ AB = PQ (Given)\\ BM = QN [From Equation (1)] \\ AM = PN (Given)\\ \Delta ABM \cong \Delta PQN (By SSS congruence rule)\\ \angle ABM = \angle PQN (By CPCT)\\ \angle ABC = \angle PQR ... (2) \\ (ii) In \Delta ABC and \Delta PQR, \\ AB = PQ (Given)\\ \angle ABC = \angle PQR [From Equation (2)]\\ BC = QR (Given)\\ \therefore \Delta ABC \cong \Delta PQR (By SAS congruence rule) 20 BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles. ##### Solution : In \Delta BEC and \Delta CFB, \\ \angle BEC = \angle CFB (Each 90^o)\\ BC = CB (Common)\\ BE = CF (Given)\\ \therefore \Delta BEC \cong \Delta CFB (By RHS congruency)\\ \therefore \angle BCE = \angle CBF (By CPCT)\\ \therefore AB = AC (Sides opposite to equal angles of a triangle are equal) \\ Hence, \Delta ABC is isosceles. 21 ABC is an isosceles triangle with AB = AC. Draw AP \perp BC to show that \angle B = \angle C. \\ ##### Solution : In \delta APB and \Delta APC, \\ \angle APB = \angle APC (Each 90^o) \\ AB =AC (Given)\\ AP = AP (Common)\\ \therefore \Delta APB \cong \Delta APC (Using RHS congruence rule)\\ \therefore \angle B = \angle C (By using CPCT) 22 Show that in a right angled triangle, the hypotenuse is the longest side. ##### Solution : Let us consider a right-angled triangle ABC, right-angled at B. \\ In \Delta ABC, \\ \angle A + \angle B + \angle C = 180^o (Angle sum property of a triangle) \\ \angle A + 90^o + \angle C = 180^o \\ \angle A + \angle C = 90^o \\ Hence, the other two angles have to be acute (i.e., less than 90^o).\\ \angle B is the largest angle in \Delta ABC. \\ \angle B > \angle A and \angle B > \angle 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 \Delta ABC. \\ However, AC is the hypotenuse of \Delta ABC. \\ Therefore, hypotenuse is the longest side in a right- angled triangle. 23 In the given figure sides AB and AC of \Delta ABC are extended to points P and Q respectively. Also, \angle PBC < \angle QCB. Show that AC > AB. ##### Solution : In the given figure,\\ \angle ABC + \angle PBC = 180^o (Linear pair)\\ \angle ABC = 180^o - \angle PBC ... (1)\\ Also,\\ \angle ACB + \angle QCB = 180^o \\ \angle ACB = 180^o -\angle QCB ... (2)\\ As \angle PBC < \angle QCB, \\ 180^o - \angle PBC > 180^o - \angle QCB \\ \angle ABC > \angle ACB [From Equations (1) and (2)]\\ AC > AB (Side opposite to the larger angle is larger.)\\ Hence proved, AC > AB 24 In the given figure, \angle B < \angle A and \angle C < \angle D. Show that AD < BC. ##### Solution : In \Delta AOB, \\ \angle B < \angle A \\ AO < BO (Side opposite to smaller angle is smaller)\\ In \Delta COD, \\ \angle C < \angle D \\ OD < OC (Side opposite to smaller angle is smaller)\\ On adding Equations (1) and (2), we obtain AO + OD < BO + OC \\ AD < BC , proved 25 AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that \angle A > \angle C and \angle B > \angle D. ##### Solution : Let us join AC.$$\\$ In $\Delta ABC,$ $\\$ $AB < BC (AB$ is the smallest side of quadrilateral $ABCD) \angle 2 < \angle 1$ (Angle opposite to the smaller side is smaller)$\\$ In $\Delta ADC,$ $\\$ $AD < CD (CD$ is the largest side of quadrilateral $ABCD) \angle 4 < \angle 3$ (Angle opposite to the smaller side is smaller)$\\$ On adding Equations (1) and (2), we obtain$\\$ $\angle 2 + \angle 4 < \angle 1 + \angle 3$ $\\$ $\angle C < \angle A4$ $\\$ $\angle A > \angle C$ $\\$ Let us join $BD$.

In $\Delta ABD,$$\\$ $AB < AD (AB$ is the smallest side of quadrilateral $ABCD)$ $\\$ $\angle 8 < \angle 5$ (Angle opposite to the smaller side is smaller)$\\$ In $\Delta BDC,$ $\\$ $BC < CD (CD$ is the largest side of quadrilateral $ABCD) \angle 7 < \angle 6$ (Angle opposite to the smaller side is smaller)$\\$ On adding Equations (3) and (4), we obtain $\\$ $\angle 8 + \angle 7 < \angle 5 + \angle 6$ $\\$ $\angle D < \angle B$ $\\$ $\angle B > \angle D$ (Hence, proved)

26   In the given figure,$PR > PQ$ and $PS$ bisects $\angle QPR$. Prove that $\angle PSR > \angle PSQ.$

##### Solution :

As $PR > PQ,$ $\\$ $\angle PQR > \angle PRQ$ (Angle opposite to larger side is larger) $\\$ $PS$ is the bisector of $\angle QPR.$ $\\$ $\angle QPS = \angle RPS ... (2)$ $\\$ $\angle PSR$ is the exterior angle of $\Delta PQS.$ $\\$ $\angle PSR = \angle PQR + \angle QPS ... (3)$ $\\$ $\angle PSQ$ is the exterior angle of $\Delta PRS.$ $\\$ $\angle PSQ = \angle PRQ + \angle RPS ... (4)$ $\\$ Adding Equations (1) and (2), we obtain $\\$ $\angle PQR + \angle QPS > \angle PRQ + \angle RPS$ $\\$ $\angle PSR > \angle PSQ$ [Using the values of Equations (3) and (4)]$\\$

27   Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

##### Solution :

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 $\Delta PNM,$ $\\$ $\angle N = 90^o$ $\\$ $\angle P + \angle N + \angle M = 180^o$ (Angle sum property of a triangle) $\\$ $\angle P + \angle M = 90^o$ $\\$ Clearly, $\angle M$ is an acute angle.$\\$ $\angle M < \angle N$ $\\$ $PN < PM$ (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.

28   $ABC$ is a triangle. Locate a point in the interior of $\Delta ABC$ which is equidistant from all the vertices of $\Delta ABC.$

##### Solution :

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 $\Delta ABC$, 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 $\Delta ABC.$

29   In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

##### Solution :

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 $\Delta ABC$, 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 $\Delta ABC.$

30   In a huge park people are concentrated at three points (see the given figure)$\\$ A: where there are different slides and swings for children,$\\$ B: near which a man-made lake is situated,$\\$ C: which is near to a large parking and exit.$\\$ Where should an ice-cream parlour be set up so that maximum number of persons can approach it?$\\$ (Hint: The parlor should be equidistant from A, B and C)

##### Solution :

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 $\Delta ABC.$ $\\$ 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.

31   Complete the hexagonal and star shaped rangolies (see the given figures) by filling them with as many equilateral triangles of side $1 cm$ as you can. Count the number of triangles in each case. Which has more triangles?

##### Solution :

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

Area of $\Delta OAB=\dfrac{\sqrt{3}}{4}(side)^2 =\dfrac{\sqrt{3}}{4}(5)^2$ $\\$ $=\dfrac{\sqrt{3}}{4}(25)=\dfrac{25\sqrt{3}}{4} cm^2$ $\\$ Area of hexagonal -shaped rangoli$=6*\dfrac{25\sqrt{3}}{4}=\dfrac{75 \sqrt{3}}{2}cm^2$ $\\$ Area of equilateral triangle having its side as $1cm=\dfrac{\sqrt{3}}{4}(1)^2=\dfrac{\sqrt{3}}{4}cm^2$ $\\$ Number of equilateral triangle of $1 cm$ side that $\\$ can be filled in this hexagonal -shaped Rangoli=$\dfrac{\dfrac{75\sqrt{3}}{2}}{\dfrac{\sqrt{3}}{4}}=150$ $\\$ Star-shaped rangoli has $12$ equilateral triangles of side $5 cm$ in it .

Area of star-shaped rangoli =$12* \dfrac{\sqrt{3}}{4}*(5)^2=75\sqrt{3}$ $\\$ Number of equilateral triangles of $1 cm$ side that can be filled in this star -shaped rangoli=$\dfrac{75\sqrt{3}}{\dfrac{\sqrt{3}}{7}}=300$ $\\$ Therefore, star-shaped rangoli has more equilateral triangles in it.