## Class 9 NCERT Maths

### NCERT

1   The angles of quadrilateral are in the ratio $3:5:9:13$. Find all the angles of the quadrilateral.

Let the common ratio between the angles be $x.$$\\ Therefore, the angles will be 3x, 5x, 9x, and 13x respectively.\\ As the sum of all interior angles of a quadrilateral is 360^o,$$\\$ $\therefore 3x+5x+9x+13x=360^o\\ 30x =360^o\\ x=12^o$$\\ Hence, the angles are\\ 3x=3*12=36^o\\ 5x=5*12=60^o\\ 9x=9*12=108^o\\ 13x=13*12=156^o 2 If the diagonals of a parallelogram are equal, then show that it is a rectangle. ##### Solution : Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is 90^o.\\ In \Delta ABC and \Delta DCB,$$\\$ $AB = DC$ (Opposite sides of a parallelogram are equal)$\\$ $BC = BC$ (Common)$\\$ $AC = DB$ (Given)$\\$ $\therefore \Delta ABC \cong \Delta DCB$ (By $SSS$ Congruence rule)$\\$ $\Rightarrow \angle ABC = \angle DCB$$\\ It is known that the sum of the measures of angles on the same side of transversal is 180^o.\\ \angle ABC + \angle DCB = 180^o (AB \parallel CD)$$\\$ $\Rightarrow ABC + \angle ABC = 180^o$$\\ \Rightarrow 2\angle ABC = 180^o$$\\$ $\Rightarrow ABC = 90^o$$\\ Since ABCD is a parallelogram and one of its interior angles is 90^o, ABCD is a rectangle. 3 Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. ##### Solution : Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e. OA = OC, OB = OD, and \angle AOB = \angle BOC = \angle COD = \angle AOD = 90^o.$$\\$ To prove $ABCD$ a rhombus,$\\$ We have to prove $ABCD$ is a parallelogram and all the sides of $ABCD$ are equal.$\\$ In $\Delta AOD$ and $\Delta COD,$$\\ OA = OC (Diagonals bisect each other)\\ \angle AOD = \angle COD (Given)\\ OD = OD (Common)\\ \therefore \Delta AOD \cong \Delta COD (By SAS congruence rule)\\ \therefore AD = CD ... (1)\\ Similarly, it can be proved that\\ AD = AB and CD = BC ... (2)\\ From Equations (1) and (2),\\ AB = BC = CD = AD$$\\$ Since opposite sides of quadrilateral $ABCD$ are equal, it can be said that $ABCD$ is a parallelogram. Since all sides of a parallelogram $ABCD$ are equal, it can be said that $ABCD$ is a rhombus.

4   Show that the diagonals of a square are equal and bisect each other at right angles.

##### Solution :

Let $ABCD$ be a square.$\\$ Let the diagonals $AC$ and $BD$ intersect each other at a point $O.$$\\ To prove that the diagonals of a square are equal and bisect each other at right angles,\\ we have to prove,\\ AC = BD, OA = OC, OB = OD, and \angle AOB = 90^o.$$\\$ In $\Delta ABC$ and $\Delta DCB$,$\\$ $AB = DC$(Sides of a square are equal to each other)$\\$ $\angle ABC = \angle DCB$ (All interior angles are of $90^o$ )$\\$ $BC = CB$ (Common side)$\\$ $\therefore \Delta ABC \cong \Delta DCB$(By $SAS$ congruency)$\\$ $\therefore AC = DB$ (By $CPCT$)$\\$ Hence, the diagonals of a square are equal in length.$\\$ In $\Delta AOB$ and $\Delta COD,$$\\ \angle AOB = \angle COD (Vertically opposite angles)\\ \angle ABO = \angle CDO (Alternate interior angles)\\ AB = CD (Sides of a square are always equal)\\ \therefore \Delta AOB \cong \Delta COD (By AAS congruence rule)\\ \therefore AO = CO and OB = OD (By CPCT)\\ Hence, the diagonals of a square bisect each other.\\ In \Delta AOB and \Delta COB,\\ As we had proved that diagonals bisect each other, therefore,\\ AO = CO$$\\$ $AB = CB$ (Sides of a square are equal)$\\$ $BO = BO$ (Common)$\\$ $\therefore \Delta AOB \cong \Delta COB$ (By $SSS$ congruency)$\\$ $\therefore \angle AOB = \angle COB$ (By $CPCT$)$\\$ However, $\angle AOB + \angle COB = 180^o$ (Linear pair)$\\$ $2\angle AOB = 180^o$$\\ \angle AOB = 90^o$$\\$ Hence, the diagonals of a square bisect each other at right angles.

5   Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

##### Solution :

Let us consider a quadrilateral $ABCD$ in which the diagonals $AC$ and $BD$ intersect each other at $O$.$\\$ It is given that the diagonals of $ABCD$ are equal and bisect each other at right angles.$\\$ Therefore, $AC = BD, OA = OC, OB = OD,$ and $\angle AOB = \angle BOC = \angle COD = \angle AOD = 90^o$.$\\$ To prove $ABCD$ is a square$\\$ We have to prove that $ABCD$ is a parallelogram, $AB = BC = CD = AD,$ and one of its interior angles is$90^o$.$\\$ In $\Delta AOB$ and $\Delta COD,$$\\ AO = CO (Diagonals bisect each other)\\ OB = OD (Diagonals bisect each other)\\ \angle AOB = \angle COD (Vertically opposite angles)\\ \therefore \Delta AOB \cong \Delta COD (SAS congruence rule)\\ \therefore AB = CD (By CPCT)$$\\$ ... (1) And, $\angle OAB = \angle OCD$(By $CPCT)$$\\ However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel.\\ AB \parallel CD ... (2)$$\\$

From Equations (1) and (2), we obtain$\\$ $ABCD$ is a parallelogram.$\\$ In $\Delta AOD$ and $\Delta COD,$$\\ AO = CO (Diagonals bisect each other)\\ \angle AOD = \angle COD (Given that each is 90^o)\\ OD = OD (Common)\\ \therefore \Delta AOD \cong \Delta COD (SAS congruence rule)\\ \therefore AD = DC ... (3)$$\\$ However,$AD = BC$ and $AB = CD$ (Opposite sides of parallelogram $ABCD$)$\\$ $AB = BC = CD = DA$$\\ Therefore, all the sides of quadrilateral ABCD are equal to each other.\\ In \Delta ADC and \Delta BCD,\\ AD = BC (Already proved)\\ AC = BD (Given)\\ DC = CD (Common)\\ \therefore \Delta ADC \cong \Delta BCD (SSS Congruence rule)\\ \therefore \angle ADC = \angle BCD (By CPCT)$$\\$ However, $\angle ADC + \angle BCD = 180^o$ (Co-interior angles)$\\$ $\angle ADC + \angle ADC = 180^o$$\\ 2\angle ADC = 180^o$$\\$ $\therefore \angle ADC = 90^o$$\\ One of the interior angles of quadrilateral ABCD is a right angle.\\ Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = ADand one of its interior angles is 90^o.$$\\$ Therefore, $ABCD$ is a square.

6   Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle A$ (see the given figure). Show that$\\$ (i) It bisects $\angle C$ also,$\\$ (ii) $ABCD$ is a rhombus.

##### Solution :

(i) $ABCD$ is a parallelogram.$\\$ $\angle DAC = \angle BCA$ (Alternate interior angles) ... (1)$\\$ And, $\angle BAC = \angle DCA$(Alternate interior angles) ... (2)$\\$ However, it is given that $AC$ bisects $\angle A.$ $\\$ $\angle DAC = \angle BAC$ ... (3)$\\$ From Equations (1), (2), and (3), we obtain$\\$ $\angle DAC = \angle BCA = \angle BAC = \angle DCA$ ... (4)$\\$ $\angle DCA = \angle BCA$ $\\$ Hence, $AC$ bisects $\angle C.$ $\\$ (ii) From Equation (4), we obtain$\\$ $\angle DAC = \angle DCA$ $\\$ $DA = DC$ (Side opposite to equal angles are equal)$\\$ However,$DA = BC$ and $AB = CD$ (Opposite sides of a parallelogram)$\\$ $AB = BC = CD = DA$ $\\$ Hence, $ABCD$ is a rhombus.

7   $ABCD$ is a rhombus. Show that diagonal $AC$ bisects $\angle A$ as well as $\angle C$ and diagonal $BD$ bisects $\angle B$ as well as $\angle D.$

##### Solution :

Let us join $AC.$ $\\$ In $\Delta ABC,$ $\\$ $BC = AB$ (Sides of a rhombus are equal to each other)$\\$ $\angle 1 = \angle 2$ (Angles opposite to equal sides of a triangle are equal)$\\$ However, $\angle 1 = \angle 3$ (Alternate interior angles for parallel lines $AB$ and $CD$)$\\$ $\angle 2 = \angle 3$ $\\$ Therefore,$AC$ bisects $\angle C.$ $\\$ Also, $\angle 2 = \angle 4$ (Alternate interior angles for || lines $BC$ and $DA$)$\\$ $\angle 1 = \angle 4$ $\\$ Therefore, $AC$ bisects $\angle A$.$\\$ Similarly, it can be proved that $BD$ bisects $\angle B$ and $\angle D$ as well.

8   $ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ as well as $\angle C$. Show that:$\\$ (i) $ABCD$ is a square$\\$ (ii) Diagonal $BD$ bisects $\angle B$ as well as $\angle D.$

##### Solution :

(i) It is given that $ABCD$ is a rectangle.$\\$ $\angle A = \angle C$ $\\$ $\rightarrow \dfrac{1}{2}\angle A =\dfrac{1}{2} \angle C$ $\\$ $\Rightarrow \angle DAC =\dfrac{1}{2}\angle DCA$ $\\$ ($AC$ bisects $\angle A$ and $\angle C$)$\\$ $CD = DA$ (Sides opposite to equal angles are also equal)$\\$ However, $DA = BC$ and $AB = CD$ (Opposite sides of a rectangle are equal)$\\$ $AB = BC = CD = DA$ $\\$ $ABCD$ is a rectangle and all the sides are equal.$\\$ Hence, $ABCD$ is a square.$\\$ (ii) Let us join $BD.$ $\\$ In $\Delta BCD,$ $\\$ $BC = CD$ (Sides of a square are equal to each other)$\\$ $\angle CDB = \angle CBD$ (Angles opposite to equal sides are equal)$\\$ However, $\angle CDB = \angle ABD$(Alternate interior angles for $AB || CD$)$\\$ $\angle CBD = \angle ABD$ $\\$ $BD$ bisects $\angle B.$ $\\$ Also, $\angle CBD = \angle ADB$ (Alternate interior angles for $BC || AD$)$\\$ $\angle CDB = \angle ABD$ $\\$ $BD$ bisects $\angle D$ and $\angle B.$

9   In parallelogram $ABCD$, two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP = BQ$ (see the given figure). Show that:$\\$ (i) $\Delta APD \cong \Delta CQB$ $\\$ (ii) $AP = CQ$ $\\$ (iii) $\Delta AQB \cong \Delta CPD$ $\\$ (iv)$AQ = CP$ $\\$ (v) $APCQ$ is a parallelogram

##### Solution :

(i) In $\Delta APD$ and $\Delta CQB,$ $\\$ $\angle ADP = \angle CBQ$ (Alternate interior angles for $BC || AD$)$\\$ $AD = CB$ (Opposite sides of parallelogram $ABCD$)$\\$ $DP = BQ$ (Given)$\\$ $\therefore APD \cong \Delta CQB$(Using $SAS$ congruence rule)$\\$ (ii) As we had observed that $\Delta APD \cong \Delta CQB,$ $\\$ $\therefore AP = CQ (CPCT)$ $\\$ (iii) In $\Delta AQB$ and $\Delta CPD,$ $\\$ $\angle ABQ = \angle CDP$(Alternate interior angles for $AB || CD$)$\\$ $AB = CD$(Opposite sides of parallelogram $ABCD$)$\\$ $BQ = DP$ (Given)$\\$ $\therefore AQB \cong \delta CPD$(Using $SAS$ congruence rule)$\\$ (iv) As we had observed that $\Delta AQB \cong \Delta CPD,$ $\\$ $\therefore AQ = CP (CPCT)$ $\\$ (v) From the result obtained in (ii) and (iv),$\\$ $AQ = CP$ and $AP = CQ$ $\\$ Since opposite sides in quadrilateral $APCQ$ are equal to each other, $APCQ$ is a parallelogram.

10   $ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$ (See the given figure). Show that $\\$ (i) $\Delta APB \cong \Delta CQD$ $\\$ (ii) $AP = CQ$ $\\$

##### Solution :

(i) In $\Delta APB$ and $\Delta CQD,$ $\\$ $\angle APB = \angle CQD$ (Each $90^o)$ $\\$ $AB = CD$(Opposite sides of parallelogram $ABCD)$ $\\$ $\angle ABP = \angle CDQ$ (Alternate interior angles for$AB || CD)$ $\\$ $\therefore \Delta APB \cong \Delta CQD$ (By $AAS$ congruency)$\\$ (ii) By using the above result$\\$ $\Delta APB \cong \Delta CQD,$ we obtain$\\$ $AP = CQ$ (By $CPCT)$ $\\$

11   In $\Delta ABC$ and $\Delta DEF, AB = DE, AB || DE, BC = EF$ and $BC || EF.$ Vertices $A, B$ and $C$ are joined to vertices $D, E$ and $F$ respectively (see the given figure). Show that$\\$ (i) Quadrilateral $ABED$ is a parallelogram$\\$ (ii) Quadrilateral $BEFC$ is a parallelogram$\\$ (iii) $AD || CF$ and $AD = CF$ $\\$ (iv) Quadrilateral $ACFD$ is a parallelogram$\\$ (v) $AC = DF$ $\\$ (vi) $\Delta ABC \cong \Delta DEF.$

##### Solution :

(i) It is given that $AB = DE$ and $AB || DE.$ $\\$ If two opposite sides of a quadrilateral are equal and parallel to each other, then it will be a parallelogram.$\\$ Therefore, quadrilateral ABED is a parallelogram.$\\$ (ii) Again,$BC = EF$ and $BC || EF$ $\\$ Therefore, quadrilateral $BCEF$ is a parallelogram.$\\$ (iii) As we had observed that $ABED$ and $BEFC$ are parallelograms, therefore$\\$ $AD = BE$ and $AD || BE$$\\ (Opposite sides of a parallelogram are equal and parallel) And, BE = CF and BE || CF$$\\$ (Opposite sides of a parallelogram are equal and parallel)$\\$ $\therefore AD = CF$ and $AD || CF$ $\\$ (iv) As we had observed that one pair of opposite sides $(AD$ and $CF)$ of quadrilateral $ACFD$ are equal and parallel to each other, therefore, it is a parallelogram.$\\$ (v) As $ACFD$ is a parallelogram, therefore, the pair of opposite sides will be equal and parallel to each other.$\\$ $\therefore AC || DF$ and $AC = DF$ $\\$ (vi) $\Delta ABC$and $\Delta DEF,$ $\\$ $AB = DE$ (Given)$\\$ $BC = EF$ (Given)$\\$ $AC = DF (ACFD$ is a parallelogram)$\\$ $\therefore ABC\cong \Delta DEF$ (By $SSS$ congruence rule)

12   $ABCD$ is a trapezium in which $AB || CD$ and $AD = BC$ (see the given figure). Show that$\\$ (i) $\angle A = \angle B$ $\\$ (ii)$\angle C = \angle D$ $\\$ (iii) $\Delta ABC \cong \Delta BAD$ $\\$ (iv) diagonal $AC$ = diagonal $BD$ $\\$ [Hint: Extend $AB$ and draw a line through $C$ parallel to $DA$ intersecting $AB$ produced at $E$.]

##### Solution :

Let us extend $AB.$ Then, draw a line through $C$, which is parallel to $AD$, intersecting $AE$ at point $E.$ It is clear that $AECD$ is a parallelogram.$\\$ (i) $AD=CE$(Opposite sides of parallelogram $AECD$)$\\$ However, $AD = BC$ (Given)$\\$ Therefore$BC = CE$ $\\$ $\angle CEB = \angle CBE$ (Angle opposite to equal sides are also equal)$\\$ Consider parallel lines $AD$ and $CE. AE$ is the transversal line for them.$\\$ $\angle A + \angle CEB = 180^o$ (Angles on the same side of transversal)$\\$ $\angle A + \angle CBE = 180^o$(Using the relation $\angle CEB = \angle CBE) ... (1)$ $\\$ However, $\angle B + \angle CBE = 180^o$(Linear pair angles) ... (2) $\\$ From Equations (1) and (2), we obtain$\\$ $\angle A = \angle B$ $\\$ (ii) $AB || CD$ $\\$ $\angle A + \angle D = 180^o$ (Angles on the same side of the transversal)$\\$ Also, $\angle C + \angle B = 180^o$ (Angles on the same side of the transversal)$\\$ $\therefore \angle A + \angle D = \angle C + \angle B$ $\\$ However, $\angle A = \angle B$[Using the result obtained in (i)]$\\$ $\therefore C = \angle D$ $\\$ (iii) In $\Delta ABC$ and $\Delta BAD$ $\\$, $AB = BA$ (Common side)$\\$ $BC = AD$ (Given)$\\$ $\angle B = \angle A$ (Proved before)$\\$ $\therefore \delta ABC \cong \Delta BAD$ (SAS congruence rule)$\\$ (iv) We had observed that,$\\$ $\Delta ABC \cong \Delta BAD$ $\\$ $\therefore AC = BD$ (By$CPCT$)

13   $ABCD$ is a quadrilateral in which $P, Q, R$ and $S$ are mid-points of the sides $AB, BC, CD$ and $DA$ (see the given figure). $AC$ is a diagonal. Show that:$\\$ (i) $SR || AC$ and $SR =\dfrac{1}{2}AC$ $\\$ (ii) $PQ = SR$ $\\$ (iii) $PQRS$ is a parallelogram.

(i) In $\Delta ADC, S$ and $R$ are the mid-points of sides $AD$ and $CD$ respectively.$\\$ In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it. $\therefore SR || AC$ and $SR =\dfrac{1}{2}AC ... (1)$ $\\$ (ii) In $\Delta ABC, P$ and $Q$ are mid-points of sides $AB$ and $BC$ respectively. Therefore, by using mid- point theorem,$\\$ $PQ || AC$ and $PQ =\dfrac{1}{2}AC$ ... (2)$\\$ Using Equations (1) and (2), we obtain$\\$ $PQ || SR$ and $PQ = SR$ ... (3) $\\$ $\therefore PQ = SR$ $\\$ (iii) From Equation (3), we obtained$\\$ $PQ || SR$ and $PQ = SR$$\\ Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal.\\ Hence, PQRS is a parallelogram. 14 ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. ##### Solution : In \Delta ABC, P and Q are the mid-points of sides AB and BC respectively. \therefore PQ || AC and PQ =\dfrac{1}{2}AC (Using mid-point theorem) ... (1)\\ In \Delta ADC, \\ R and S are the mid-points of CD and AD respectively.\\ \therefore RS || AC and RS =\dfrac{1}{2}AC (Using mid-point theorem) ... (2)\\ From Equations (1) and (2), we obtain\\ PQ || RS and PQ = RS$$\\$ Since in quadrilateral $PQRS,$ one pair of opposite sides is equal and parallel to each other, it is a parallelogram.$\\$ Let the diagonals of rhombus $ABCD$ intersect each other at point $O.$ In quadrilateral $OMQN,$ $\\$ $MQ||ON(\therefore PQ||AC)$$\\ QN || OM (\therefore QR || BD) \\ Therefore, OMQN is a parallelogram.\\ \therefore MQN = \angle NOM$$\\$ $\therefore \angle PQR =\angle NOM$ $\\$ However$\angle NOM = 90^o$ (Diagonals of a rhombus are perpendicular to each other)$\\$ $\therefore PQR = 90^o$ $\\$ Clearly, $PQRS$ is a parallelogram having one of its interior angles as$90^o.$ $\\$ Hence, $PQRS$ is a rectangle.

15   $ABCD$ is a rectangle and $P, Q, R$ and $S$ are mid-points of the sides $AB, BC, CD$ and $DA$ respectively. Show that the quadrilateral$PQRS$ is a rhombus.

##### Solution :

Let us join $AC$ and $BD.$ $\\$ In $\Delta ABC,$ $\\$ $P$ and $Q$ are the mid-points of $AB$ and $BC$ respectively.$\\$ $\therefore PQ || AC$ and $PQ =\dfrac{1}{2}AC$(Mid-point theorem) ... (1)$\\$ Similarly in $\Delta ADC,$ $\\$ $SR || AC$ and $SR =\dfrac{1}{2}AC$ (Mid-point theorem) ... (2)$\\$ Clearly, $PQ || SR$ and $PQ = SR$ $\\$ Since in quadrilateral $PQRS,$ one pair of opposite sides is equal and parallel to each other, it is a parallelogram.$\\$ $\therefore PS || QR$ and $PS = QR$ (Opposite sides of parallelogram) ... (3)$\\$ In $\Delta BCD, Q$ and $R$ are the mid-points of side $BC$ and $CD$ respectively.$\\$ $\therefore QR || BD$ and $QR =\dfrac{1}{2}BD$ (Mid-point theorem) ... (4)$\\$ However, the diagonals of a rectangle are equal.$\\$ $\therefore AC = BD ...(5)$ $\\$ By using Equations (1), (2), (3), (4), and (5), we obtain$\\$ $PQ = QR = SR = PS$ $\\$ Therefore,$PQRS$ is a rhombus.

16   $ABCD$ is a trapezium in which $AB || DC, BD$ is a diagonal and $E$ is the mid - point of $AD. A$ line is drawn through $E$ parallel to $AB$ intersecting $BC$ at $F$ (see the given figure). Show that $F$ is the mid-point of $BC.$

Let $EF$ intersect $DB$ at $G.$$\\ By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side.\\ In \Delta ABD, \\ EF || AB and E is the mid-point of AD. \\ Therefore, G will be the mid-point of DB. \\ As EF || AB and AB || CD, \\ \therefore EF || CD (Two lines parallel to the same line are parallel to each other)\\ In \Delta BCD, GF || CD and G is the mid-point of line BD. \\Therefore, by using converse of mid-point theorem, F is the mid-point of BC. 17 In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD. ##### Solution : ABCD is a parallelogram.\\ AB || CD \\ And hence, AE || FC$$\\$ Again, $AB = CD$ (Opposite sides of parallelogram $ABCD$)$\\$ $\dfrac{1}{2}AB =\dfrac{1}{2}CD$ $\\$ $AE = FC$($E$ and$F$ are mid-points of side $AB$ and $CD$)$\\$ In quadrilateral $AECF$, one pair of opposite sides ($AE$ and $CF$) is parallel and equal to each other. Therefore, $AECF$ is a parallelogram.$\\$ $\therefore AF || EC$ (Opposite sides of a parallelogram)$\\$ In $\Delta DQC, F$ is the mid-point of side $DC$ and $FP || CQ$ (as $AF || EC$). Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of $DQ.$$\\ \therefore DP = PQ ... (1)\\ Similarly, in \Delta APB, E is the mid-point of side AB and EQ || AP (as AF || EC).\\ Therefore, by using the converse of mid-point theorem, it can be said that Q is the mid-point of PB.$$\\$ $\therefore PQ = QB$ ... (2)$\\$ From Equations (1) and (2),$\\$ $DP = PQ = BQ$ $\\$ Hence, the line segments $AF$ and $EC$ trisect the diagonal $BD.$

18   Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

##### Solution :

Let $ABCD$ is a quadrilateral in which $P, Q, R$, and $S$ are the mid-points of sides $AB, BC, CD,$ and$DA$ respectively. Join $PQ, QR, RS, SP,$ and$BD.$ $\\$ In $\Delta ABD, S$ and $P$ are the mid-points of $AD$ and $AB$respectively. Therefore, by using mid-point theorem, it can be said that $\\$ $SP || BD$ and $SP =\dfrac{1}{2}BD ... (1)$ $\\$ Similarly in $\Delta BCD,$ $\\$ $QR || BD$ and $QR =\dfrac{1}{2}BD$ ... (2)$\\$ From Equations (1) and (2), we obtain$\\$ $SP || QR$ and $SP = QR$ $\\$ In quadrilateral $SPQR,$ one pair of opposite sides is equal and parallel to each other. Therefore, $SPQR$ is a parallelogram.$\\$ We know that diagonals of a parallelogram bisect each other.$\\$ Hence,$PR$ and $QS$ bisect each other.

19   $ABC$ is a triangle right angled at $C. A$ line through the mid-point $M$ of hypotenuse $AB$ and parallel to $BC$ intersects $AC$ at $D$. Show that$\\$ (i) $D$ is the mid-point of $AC$ $\\$ (ii) $MD \perp AC$ $\\$ (iii) $CM = MA =\dfrac{1}{2} AB$

##### Solution :

(i) In $\Delta ABC,$ $\\$ It is given that M is the mid-point of $AB$ and $MD || BC.$ $\\$ Therefore, $D$ is the mid-point of $AC.$ (Converse of mid-point theorem)$\\$ (ii) As $DM || CB$ and $AC$ is a transversal line for them, therefore,$\\$ $\angle MDC +\angle DCB = 180^o$ (Co-interior angles)$\\$ $\angle MDC + 90^o = 180^o$$\\ \angle MDC = 90^o \\ \therefore MD \perp AC \\ (iii) Join MC.$$\\$ In $\Delta AMD$ and $\Delta CMD,$ $\\$ $AD = CD$ ($D$ is the mid-point of side $AC$)$\\$ $\angle ADM = \angle CDM$(Each $90^o$)$\\$ $DM = DM$(Common)$\\$ $\therefore AMD \cong \Delta CMD$ (By $SAS$ congruence rule)$\\$ Therefore, $AM = CM$(By $CPCT$)$\\$ However, $AM =\dfrac{1}{2}AB$($M$ is the mid-point of $AB$)$\\$ Therefore, it can be said that$\\$ $CM = AM =\dfrac{1}{2}AB$