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 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) $\angle ABD = \angle BAC.$

Solution :

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 $AB$.

Solution :

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.$$\\$

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 :

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 $BC = DE.$

Solution :

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) $AD = BE$

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$

Solution :

(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.