Quadrilaterals

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.

Solution :

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$$\\$ One of the interior angles of quadrilateral $ABCD $ is a right angle.$\\$ Thus, we have obtained that $ABCD$ is a parallelogram, $AB = BC = CD = AD$and one of its interior angles is $90^o.$$\\$ 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.

Solution :

(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 $BC.$

Solution :

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$\\$ $CM = AM =\dfrac{1}{2}AB$