# Question-1

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

**Solution:**

Let ABCD be a quadrilateral in which âˆ A : âˆ B : âˆ C :âˆ D = 3 : 5 : 9 : 13

Sum of the ratios = 3 + 5 + 9 + 13 = 30

Also, âˆ A + âˆ B + âˆ C + âˆ D = 360Â°

Sum of all the angles of a quadrilateral is 360Â°

âˆ´ âˆ A = Ã— 360Â° = 36Â°

âˆ B = Ã— 360Â° = 60Â°

âˆ C = Ã— 360Â° = 108Â°

and âˆ D = Ã— 360Â° = 156Â°

# Question-2

**If the diagonals of a parallelogram are equal, then show that it is a rectangle.**

**Solution:**

**Given:**In parallelogram ABCD, AC = BD

**To Prove: ** Parallelogram ABCD is a rectangle.

**Proof:** In Î” ACB and Î” BDA,

AC = BD | Given

AB = BA | Common

BC = AD |Opposite sides of the parallelogram ABCD

D ACB â‰… Î” BDA | SSS Rule

âˆ´ âˆ ABC = âˆ BAD ...(1) CPCT

Again AD || BC | Opposite sides of parallelogram ABCD

AD || BC and the transversal AB intersects them.

âˆ´ âˆ BAD + âˆ ABC = 180Â° ...(2) |Sum of consecutive interior angles on |the same side of the transversal is 180Â°

From (1) and (2),

âˆ BAD = âˆ ABC = 90Â°

âˆ´ âˆ A = 90Â° and âˆ C = 90Â°

Similarly we can prove âˆ B = 90Â° and âˆ D = 90Â°

Parallelogram ABCD is a rectangle.

# Question-3

**Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.**

**Solution:**

**Given:**ABCD is a quadrilateral where diagonals AC and BD intersect each other at right angles at O.

**To Prove: **Quadrilateral ABCD is a rhombus.

**Proof: ** In Î” AOB and Î” AOD,

AO = AO | Common

OB = OD | Given âˆ AOB = âˆ AOD | Each = 90Â°

âˆ´ Î” AOB â‰… Î” AOD | SAS Rule

AB = AD ...(1) | CPCT

Similarly, we can prove that

AB = BC ...(2)

BC = CD ...(3)

CD = AD ...(4)

From (1), (2), (3) and (4), we obtain

AB = BC = CD = DA âˆ´ Quadrilateral ABCD is a rhombus.

# Question-4

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

**Solution:**

**Given:**ABCD is a square.

**To Prove:**(i) AC = BD

(ii) AC and BD bisect each other at right angles.

**Proof:**(i) In Î” ABC and Î” BAD,

AB = BA | Common

BC = AD | Opposite sides of square ABCD

âˆ ABC = âˆ BAD | Each = 90Â° (ABCD is a square)

Î” ABC â‰… Î” BAD | SAS Rule

AC = BD | CPCT

(ii) In Î” OAD and Î” OCB,

AD = CB | Opposite sides of square ABCD

âˆ OAD = âˆ OCB | AD || BC and transversal AC intersects them âˆ ODA = âˆ OBC | AD || BC and transversal BD intersects them Î” OAD â‰… Î” OCB | ASA Rule

âˆ´ OA = OC â€¦..(1)

Similarly, we can prove that

OB = OD â€¦..(2)

From (1) and (2),

AC and BD bisect each other.

Again, in Î” OBA and Î” ODA,

OB = OD | From (2) above

BA = DA | Sides of square are equal

OA = OA | Common

âˆ´ Î” OBA â‰… Î” ODA | SSS Rule

âˆ´ âˆ AOB = âˆ AOD â€¦..(3) | CPCT

But âˆ AOB + âˆ AOD = 180Â° | Linear Pair Axiom

âˆ´ âˆ AOB = âˆ AOD = 90Â° | From (3)

âˆ´ AC and BD bisect each other at right angles.

# Question-5

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

**Solution:**

**Given:**The diagonals AC and BD of a quadrilateral ABCD and equal and bisect each other at right angles.

**To Prove: ** Quadrilateral ABCD is a square.

**Proof: **In Î” OAD and Î” OCB,

OA = OC | Given

OD = OB | Given âˆ AOD = âˆ COB | Vertically Opposite Angles Î” OAD â‰… Î” OCB | SAS Rule AD = CB | c.p.c.t. âˆ ODA = âˆ OBC | c.p.c.t. âˆ´âˆ ODA = âˆ OBC

âˆ´ AD || BC

Now, AD = CB and AD || CB

âˆ´ Quadrilateral ABCD is a || gm.

Similarly we can prove that AB = DC and AB || DC

In Î” AOB and Î” AOD,

AO = AO | Common

OB = OD | Given â‡’ âˆ AOB = âˆ AOD | Each = 90Â° (Given) â‡’ Î” AOB â‰… Î” AOD | SAS Rule â‡’ AB = AD

Now, ABCD is a parallelogram and AB = AD

âˆ´ ABCD is a rhombus.

Again, in Î” ABC and Î” BAD,

AC = BD | Given

BC = AD |

âˆ´ ABCD is a rhombus

AB = BA | Common â‡’ Î” ABC â‰… Î” BAD | SSS Rule â‡’ âˆ ABC = âˆ BAD | c.p.c.t. â‡’ AD || BC | Opp. sides of || gm ABCD and transversal AB intersects them.

â‡’ âˆ ABC + âˆ BAD = 180Â° | Sum of consecutive interior angles on the same side of the transversal is 180Â°

âˆ´ âˆ ABC = âˆ BAD = 90Â°

Similarly, âˆ BCD = âˆ ADC = 90Â°

âˆ´ ABCD is a square.

# Question-6

**Diagonal AC of a parallelogram ABCD bisects**âˆ

**A (see figure). Show that:**

**(i) it bisects**âˆ

**C also**

**(ii) ABCD is a rhombus.**

**Solution:**

**Given:**Diagonal AC of a parallelogram ABCD bisects âˆ A.

**To Prove:**(i) it bisects âˆ C also.

(ii) ABCD is a rhombus.

**Proof:**(i) In Î” ADC and Î” CBA,

AD = CB | Opp. sides of parallelogram ABCD

CA = CA | Common

DC = BA | Opp. sides of parallelogram ABCD

âˆ´ Î” ADC â‰… Î” CBA | SSS Rule

âˆ´ âˆ ACD = âˆ CAB | c.p.c.t.

and âˆ DAC = âˆ BCA | c.p.c.t.

But âˆ CAB = âˆ DAC | Given

\ âˆ ACD = âˆ BCA

âˆ´ AC bisects âˆ C also.

(ii) From above, âˆ ACD = âˆ CAD

\ AD = CD | Opposite sides of equal angles of a triangle are equal

\ AB = BC = CD = DA |

âˆ´ ABCD is a parallelogram

ABCD is a rhombus.

# Question-7

**ABCD is a rhombus. Show that diagonal AC bisects**âˆ

**A as well as**âˆ

**C and diagonal BD bisects**âˆ

**B as well as**âˆ

**D.**

**Solution:**

**Given :**ABCD is a rhombus.

**To Prove :**(i) Diagonal AC bisects âˆ A as well as âˆ C.

(ii) Diagonal BD bisects âˆ B as well as âˆ D.

**Proof:**(i)

**ABCD is a rhombus**

AD = CD

âˆ´ âˆ DAC = âˆ DCA ...(1) | Angles opposite to equal sides of a triangle are equal

Also, CD || AB and transversal AC intersects them

âˆ´ âˆ DAC = âˆ BCA â€¦(2) |Alternate Interior angles

From (1) and (2) âˆ DCA = âˆ BCA

â‡’ AC bisects âˆ C

Similarly AC bisects âˆ A.

(ii) Proceeding similarly as in (i) above, we can prove that BD bisects âˆ B as well as âˆ D.

# Question-8

**ABCD is a rectangle in which diagonal AC bisects**âˆ

**A as well as**âˆ

**C. Show that (i) ABCD is a square (ii) diagonal BD bisects**âˆ

**B as well as**âˆ

**D.**

**Solution:**

**Given:**ABCD is a rectangle in which diagonal AC bisects âˆ A as well as âˆ C.

**To Prove:** (i) ABCD is a square.

(ii) diagonal BD bisects âˆ B as well as âˆ D.

**Proof: ** (i) AB || DC and transversal AC intersects them.

âˆ´ âˆ ACD = âˆ CAB |Alternate interior angles

But âˆ CAB = âˆ CAD

âˆ´ âˆ ACD = âˆ CAD

âˆ´ AD = CD |Sides opposite to equal angles of a triangle are equal

âˆ´ ABCD is a square.

(ii) In Î” BDA and Î” DBC,

BD = DB | Common

DA = BC | Sides of a square ABCD

AB = DC | Sides of a square ABCD

âˆ´ Î” BDA â‰… Î” DBC | SSS Rule

âˆ´ âˆ ABD = âˆ CDB | CPCT

But âˆ CDB = âˆ CBD | âˆ´ CB = CD (Sides of a square ABCD)

âˆ´ âˆ ABD = âˆ CBD .......... (1)

âˆ´ BD bisects âˆ B.

Now, âˆ ABD = âˆ CBD | from (1)

âˆ ABD = âˆ ADB |âˆ´ AB = AD

âˆ CBD = âˆ CDB |âˆ´ CB = CD

âˆ CDB = âˆ CBD = âˆ ABD = âˆ ADB

âˆ CDB = âˆ ADB âˆ´ BD bisects âˆ D.

# Question-9

**In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that:**

(i) Î” APD â‰… Î” CQB

(ii) AP = CQ

(iii) Î” AQB â‰… Î” CPD

(iv) AQ= CP

(v) APCQ is a parallelogram

(i) Î” APD â‰… Î” CQB

(ii) AP = CQ

(iii) Î” AQB â‰… Î” CPD

(iv) AQ= CP

(v) APCQ is a parallelogram

**Solution:**

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**Given:**In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ.

**To Prove: ** (i) Î” APD â‰… Î” CQB

(ii) AP = CQ

(iii) Î” AQB = Î” CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

**Construction: ** Join AC to intersect BD at O.

**Proof: **

First we need to prove APCQ is a parallelogram

(v) The diagonals of a parallelogram bisect each other.

â‡’ OB = OD

â‡’ OB - BQ = OD - DP | BQ = DP given

â‡’ OQ = OP .......... (1)

Also, OA = OC .......... (2) | Diagonals of a parallelogram bisect each other

From (1) and (2), APCQ is a parallelogram.

(ii) APCQ is a parallelogram | Proved in (v) above âˆ´ AP = CQ | Opposite sides of a parallelogram are equal

(iv) âˆ´ APCQ is a parallelogram| Proved in (i) above

âˆ´ AQ = CP| Opposite sides of a parallelogram are equal

(i) In Î” APD and Î” CQB,

AP = CQ | Opposite sides of a parallelogram are equal | from (ii)

PD = QB | Given

AD = CB| Opposite sides of a parallelogram are equal

âˆ´ Î” APD â‰… Î” CQB.| SSS rule

(iii) In Î” AQB and Î” CPD,

AQ = CP | APCQ is a parallelogram

QB = PD | Given

AB = CD | Opposite sides of parallelogram ABCD

âˆ´ AQB â‰… CPD | SSS Rule

# Question-10

**ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively (see figure).**

**Show that:**

(i)

(i)

**D**

**APB**â‰… Î”

**CQD**

(ii) AP = CQ.

(ii) AP = CQ.

**Solution:**

**Given:**ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively.

**To Prove:**(i) Î” APB â‰… Î” CQD

(ii) AP = CQ.

**Proof:**(i) In Î” APB and Î” CQD,

AB = CD | Opposite sides of parallelogram ABCD

âˆ ABP = âˆ CDQ |

âˆ´ AB || DC and transversal BD intersects them âˆ APB = âˆ CQD | Each = 90Â°

âˆ´ âˆ APB â‰… âˆ CQD | AAS Rule

(ii) âˆ´ Î” APB â‰… Î” CQD | Proved above in

(i) âˆ´ AP = CQ. | CPCT

# Question-11

**In**

**D**

**ABC and**Î”

**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 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) Î” ABC****â‰… Î” DEF**

**Solution:**

**Given:**In Î” ABC and Î” DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively.

**To Prove:**(i) Quadrilateral ABED is a parallelogram

(ii) Quadrilateral BECF is a parallelogram

(iii) AD || CF and AD = CF

(iv) Quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) Î” ABC â‰… Î” DEF

**Proof:**(i) In quadrilateral ABED,

AB = DE and AB || DE | Given

âˆ´ Quadrilateral ABED is a parallelogram

(ii) In quadrilateral BEFC,

BC = EF and BC || EF | Given

âˆ´ Quadrilateral BECF is a parallelogram

(iii) ABED is a parallelogram | From (i)

âˆ´ AD || BE and AD = BE ...(1) |

**âˆµ**Opposite sides of a parallelogram are parallel and equal

BEFC is a parallelogram | From (ii)

âˆ´ BE || CF and BE = CF ...(2) |

**âˆµ**Opposite sides of a parallelogram are parallel and equal

From (1) and (2), we obtains AD || CF and AD = CF

(iv) In quadrilateral ACFD,

AD || CF and AD = CF | Proved in (iii)

âˆ´ Quadrilateral ACFD is a parallelogram

(v) ACFD is a parallelogram | Proved in (iv)

AC || DF and AC = DF |

**âˆµ**In a parallelogram opposite sides are parallel and of equal length

(vi) In Î” ABC and Î” DEF

AB = DE |

**âˆµ**ABED is a parallelogram

BC = EF |

**âˆµ**BEFC is a parallelogram

AC = DF | Proved in (v)

âˆ´ Î” ABC â‰… Î” DEF | SSS Rule

# Question-12

**ABCD is a trapezium in which AB ||**

**CD and AD = BC (see figure). Show that:**

**(i)**âˆ

**A =**âˆ

**B**

**(ii)**âˆ

**C =**âˆ

**D**

**(iii)**Î”

**ABC**â‰… Î”

**BAD**

**(iv) Diagonal AC = Diagonal BD**

**[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E]**

**Solution:**

**Given:**ABCD is a trapezium in which AB || CD and AD = BC

**To Prove:**(i) âˆ A = âˆ B

(ii) âˆ C = âˆ D

(iii) Î” ABC â‰… Î” BAD

(iv) Diagonal AC = Diagonal BD

**Construction:**Extend AB and draw a line through C parallel to DA intersecting AB produced at E.

**Proof:**

(i) AB || CD | Given and AD || EC | By construction

âˆ´ AECD is a parallelogram

âˆ´ AD = EC | Opposite sides of a parallelogram are equal

But AD = BC | Given

EC = BC

âˆ´ âˆ CBE = âˆ CEB ...(1) | Angles of opposite to equal sides of a triangle are equal âˆ B + âˆ CBE = 180Â° ...(2) | Linear Pair Axiom âˆ A +âˆ CEB = 180Â°....(3) | The sum of consecutive interior angles on the same side of the transversal is 180Â°

From (2) and (3), âˆ B + âˆ CBE = âˆ A + âˆ CEB

But âˆ CBE = âˆ CEB

âˆ´ âˆ B = âˆ A or âˆ A = âˆ B

(ii) AB || CD âˆ A + âˆ D = 180Â°

and âˆ B + âˆ C = 180Â°

âˆ´ âˆ A + âˆ D = âˆ B + âˆ C | Proved in (i)

But âˆ A = âˆ B

âˆ´ âˆ D = âˆ C or âˆ C = âˆ D

(iii) In Î” ABC and Î” BAD, AB = BA | Common

BC = AD | Given âˆ ABC = âˆ BAD | From (i)

âˆ´ Î” ABC â‰… Î” BAD | SAS Rule

(iv) Î” ABC = Î” BAD | From (iii) above

âˆ´ AC = BD | CPCT

# Question-13

**ABCD is a quadrilateral in which p, Q, R and S are mid-points of the sides AB, BC, CD and DA, AC is a diagonal. Show that:**

**(i) SR || AC and SR =**

**AC**

**(ii) PQ = SR**

**(iii) PQRS is a parallelogram.**

**Solution:**

**Given:**ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal.

**To Prove:**(i) SR || AC and SR = AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

**Proof:**(i) In Î” DAC,

âˆ´ S is the mid-point of DA and R is the mid-point of DC

âˆ´ SR || AC and SR = AC |Mid-point theorem

(ii) In Î” BAC,

âˆ´ P is the mid-point of AB and Q is the mid-point of BC

âˆ´ PQ || AC and PQ = AC | Mid-point theorem

But from (i), SR = AC âˆ´ PQ = SR

(iii) PQ || AC | From (ii)

SR || AC | From(i) âˆ´ PQ || SR | Two lines parallel to the same line are parallel to each other

Also, PQ = SR | From (ii)

âˆ´ PQRS is a parallelogram.

# Question-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:**

**Given:**ABCD is a rhombus. P, Q, R, S are the mid - points of AB, BC, CD, DA respectively. PQ, QR, RS and SP are joined.

**To prove:**PQRS is a rectangle.

**Construction:**Join AC and BD.

**Proof:**In triangles RDS and PBQ,

DS = QB |âˆµ Halves of opposite sides of parallelogram ABCD which are equal

DR = PB |âˆµ Halves of opposite sides of parallelogram ABCD which are equal

âˆ SDR = âˆ QBP |âˆµ Opposite angles of parallelogram ABCD which are equal

âˆ´ Î” RDS

**â‰…**Î” PBQ |âˆµ SAS Axiom

âˆ´ SR = PQ |âˆµ c.p.c.t.

In triangles RCQ and PAS,

RC = AP |âˆµ Halves of opposite sides of parallelogram ABCD which are equal

CQ = AS |âˆµ Halves of opposite sides of parallelogram ABCD which are equal

âˆ RCQ = âˆ PAS | âˆµ Opposite angles of parallelogram ABCD which are equal

âˆ´ Î” RCQ = Î” PAS

RQ = SP

âˆ´ In PQRS,

SR = PQ and RQ = SP âˆ´ PQRS is a parallelogram,

In Î” CDB, âˆ´ R and Q are the mid-points of DC and CB respectively.

RQ || DB â‡’ RF || EO.

Similarly, RE || FO

âˆ´ OFRE is a parallelogram

âˆ´ âˆ R = âˆ EOF = 90Â° |

âˆµ Opposite angles of a parallelogram are equal and diagonals of a rhombus intersect at 90Â°

Thus PQRS is a rectangle.

# Question-15

**ABCD is a rectangle 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 rhombus.**

**Solution:**

**Given:**ABCD is a rectangle. P, Q, R and S are mid-points of AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.

**To prove:**Quadrilateral PQRS is a rhombus.

**Construction:**Join AC.

**Proof: **In Î” ABC,

âˆ´ P and Q are the mid-points of AB and BC respectively.

âˆ´ PQ || AC and PQ = AC â€¦..(1)

In Î” ADC,

âˆ´ S and R are the mid-points of AD and DC respectively.

âˆ´ SR || AC and SR = AC â€¦..(2)

From (1) and (2),

PQ || SR and PQ = SR

âˆ´ Quadrilateral PQRS is a parallelogram â€¦.(3)

In rectangle ABCD,

AD = BC | Opposite sides â‡’ AD = BC |

âˆµ Halves of equals are equal

â‡’ AS = BQ

In Î” APS and Î” BPQ,

AP= BP |

âˆµ P is the mid-point of AB

AS = BQ | Proved above âˆ PAS = âˆ PBQ

| Each = 90Â° âˆ´ Î” APS â‰… Î” BPQ | SAS Axiom

âˆ´ PS = PQ ...(4) | c.p.c.t.

In view of (3) and (4), PQRS is a rhombus.

# Question-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 figure). Show that F is the mid-point of BC.**

**Solution:**

**Given:**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 R

**To Prove: ** F is the mid-point of BC.

**Proof: ** Let DB intersect EF at G.

In Î” DAB, E is the mid-point of DA and EG || AB

âˆ´ G is the mid-point of DB | By converse of mid-point theorem

Again, in Î” BDC, G is the mid-point of BD and GF ||AB || DC

âˆ´ F is the mid-point of BC. | By converse of mid-point theorem

# Question-17

**In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively, (see figure). Show that the line segments AF and EC trisect the diagonal BD.**

**Solution:**

**Given:**In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively.

**To Prove:**Line segments AF and EC trisect the diagonal BD.

**Proof:**AB || DC ( âˆµ Opposite sides of parallelogram ABCD )

âˆ´ AE || FC ...............(1)

âˆ´ AB = DC (âˆµ Opposite sides of parallelogram ABCD )

AB = DC ( Halves of equals are equal ) â‡’ AE = CF ...............(2)

In view of (1) and (2),

AECF is a parallelogram

âˆ´ EC || AF ...........(3) ( Opposite sides of parallelogram AECF )

In Î” DBC, F is the mid-point of DC

and FP || CQ ( âˆµ EC || AF)

âˆ´ P is the mid-point of DQ ( By converse of mid-point theorem) â‡’ DP = PQ ............(4)

Similarly, in Î” BAP,

BQ = PQ ...........(5)

From (4) and (5), we obtain

DP = PQ = BQ

â‡’ Line segments AF and EC trisect the diagonal BD.

# Question-18

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

**Solution:**

**Given:**ABCD is a quadrilateral. P, Q, R and S are the mid-points of the sides DC, CB, BA and AD respectively.

**To prove:**PR and QS bisect each other.

**Construction:**Join PQ, QR, RS, SP, AC and BD.

**Proof:**In Î” ABC,

âˆ´ R and Q are the mid-points of AB and BC respectively.

âˆ´ RQ || AC and RQ = AC.

Similarly, we can show that

PS || AC and PS = AC

âˆ´ RQ || PS and RQ = PS.

Thus a pair of opposite sides of a quadrilateral PQRS are parallel and equal.

âˆ´ PQRS is a parallelogram.

Since the diagonals of a parallelogram bisect each other.

âˆ´ PR and QS bisect each other.

# Question-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**âŠ¥

**AC**

**(iii) CM = MA=**

**AB**

**Solution:**

**Given:**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.

**To Prove:**(i) D is the mid-point of AC.

(ii) MD âŠ¥ AC

(iii) CM = MA = AB.

**Proof:**(i) In Î” ACB, M is the mid-point of AB and MD || BC âˆ´ D is the mid-point of AC | By converse of mid-point theorem

(ii) MD || BC and AC intersects them

âˆ´ âˆ ADM = âˆ ACB | Corresponding angles

But âˆ ACB = 90Â° | Given

âˆ´ âˆ ADM = 90Â° â‡’ MD âŠ¥ AC

(iii) Now, âˆ ADM + âˆ CDM = 180Â° | Linear Pair Axiom

âˆ´ âˆ ADM = âˆ CDM = 90Â°

In Î” ADM and Î” CDM,

AD = CD |

âˆ´ D is the mid-point of AC âˆ ADM = âˆ CDM | Each = 90Â°

DM = DM | Common

âˆ´ Î” ADM â‰… Î” CDM | SAS Rule

âˆ´ MA = MC | CPCT

But M is the mid-point of AB

âˆ´ MA = MB = âˆ´ MA = MC = â‡’ CM = MA =