**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 $BD$?**

In $\Delta ABC$ and $\Delta ABD,$$\\$ $AC = AD$ (Given)$\\$ $\angle CAB = \angle DAB (AB$ bisects $\angle A)$$\\$ $AB = AB $(Common)$\\$ $\therefore \Delta ABC \cong \Delta ABD$ (By $SAS$ congruence rule)$\\$ $\therefore BC = BD (By \ CPCT)$$\\$ Therefore, $BC$ and $BD$ are of equal lengths.

**2** **$ABCD $ is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ (See the given figure). Prove that$\\$ (i) $\Delta ABD \cong \Delta BAC$$\\$ (ii)$ BD = AC$$\\$ (iii) $\angle ABD = \angle BAC.$**

In $\Delta ABD$ and $\Delta BAC,$$\\$ $AD = BC$ (Given)$\\$ $\angle DAB = \angle CBA $(Given)$\\$ $AB = BA$ (Common)$\\$ $\therefore \Delta ABD \cong \Delta BAC$ (By $SAS$ congruence rule)$\\$ $\therefore BD=Ac(By \ CPCT)$$\\$ And, $\angle ABD=\angle BAC(By\ CPCT) $

**3** **$AD$ and $BC$ are equal perpendiculars to a line segment $AB$ (See the given figure). Show that $CD$ bisects $AB$.**

In $\Delta BOC $ and $\Delta AOD,$$\\$ $\angle BOC = \angle AOD $(Vertically opposite angles)$\\$ $\angle CBO = \angle DAO$ (Each $90^o$)$\\$ $BC = AD$ (Given)$\\$ $\therefore \Delta BOC \cong \Delta AOD (AAS$ congruence rule)$\\$ $\therefore BO = AO (By \ CPCT)$$\\$ $ CD$ bisects $AB.$

**4** **$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see the given figure). Show that $\Delta ABC \cong \Delta CDA.$$\\$**

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 $\angle A$.**

In $\Delta APB$ and $\Delta AQB,$$\\$ $\angle APB = \angle AQB$ (Each $90^o)$$\\$ $\angle PAB = \angle QAB$ ($l$ is the angle bisector of $\angle A$)$\\$ $AB = AB$ (Common)$\\$ $\therefore \Delta APB \cong \Delta AQB$ (By $AAS$ congruence rule)$\\$ $\therefore BP = BQ$ (By $ CPCT)$$\\$ Or, it can be said that B is equidistant from the arms of $\angle A$.

**6** **In the given figure, $AC = AE, AB = AD$ and $\angle BAD = \angle EAC.$ Show that $BC = DE.$**

It is given that $\angle BAD = \angle EAC$ $\\$ $\angle BAD + \angle DAC = \angle EAC + \angle DAC$ $\\$ $\angle BAC = \angle DAE$ $\\$ In$ \Delta BAC $ and $\Delta DAE,$ $\\$ $AB = AD $ (Given)$\\$ $\angle BAC = \angle DAE $ (Proved above)$\\$ $AC = AE$ (Given)$\\$ $ \therefore \Delta BAC \cong \Delta DAE$ (By $SAS$ congruence rule)$\\$ $\therefore BC = DE $(By $CPCT$)

**7** **$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE $ and $\angle EPA = \angle DPB $ (See the given figure). Show that $\\$ (i) $\Delta DAP \cong \Delta EBP$$\\$ (ii) $AD = BE$**

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 $\angle A$**

(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 $AB = AC.$**

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.**

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.**

(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 $\angle ABD = \angle ACD.$**

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.**

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 $\angle C.$**

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.**

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 $BC.$**

(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 $\angle A.$**

(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) $\Delta ABC \cong \Delta PQR$**

(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.**

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 $\angle B = \angle C.$ $\\$**

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.**

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 $AC > AB.$**

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 $AD < BC.$**

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 $\angle B > \angle D.$**

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 $\angle PSR > \angle PSQ.$**

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.**

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 $\Delta ABC.$**

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.**

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)**

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?**

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.