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

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Given: The ratios of the angles of quadrilateral are 3: 5: 9: 13.

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.

We know that, sum of angles of a quadrilateral = \(360^\circ\)

i.e., 3x + 5x + 9x + 13x = \(360^\circ\)

i.e., 30x = \(360^\circ\)

Therefore, x = \(12^\circ\)

Thus, we get,

angles of quadrilateral

=> 3x = 3 × \(12^\circ\) = \(36^\circ\)

=>5x = 5 × \(12^\circ\) = \(60^\circ\)

=> 9x = 9 × \(12^\circ\) = \(108^\circ\)

=> 13x = 13 × \(12^\circ\) = \(156^\circ\)

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

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Given: Parallelogram is ABCD whose diagonals AC and BD are equal.

i.e., AC = BD

To prove: ABCD is a rectangle.

Proof: In \(\triangle{ABC}\) and \(\triangle{DCB}\), we have,

AC = BD ...(Given)

AB = CD ...(Opposite sides of parallelogram)

BC = CB ...(Common sides)

Therefore, \(\triangle{ABC}\)\(\displaystyle \cong \)\(\triangle{DCB}\) ...(By SSS rule)

Thus, \(\angle{ABC}\) = \(\angle{DCB}\) ...(i)(By CPCT)

But from figure, DC || AB and transversal CB intersects them

\(\angle{ABC}\) + \(\angle{DCB}\) = \(180^\circ\) ...(interior angles on same side of transversal)

\(\angle{ABC}\) + \(\angle{ABC}\) = \(180^\circ\) ...(from(i))

i.e., 2 \(\angle{ABC}\) = \(180^\circ\)

i.e., \(\angle{ABC}\) = \(90^\circ\)

Also, \(\angle{DCB}\) = \(90^\circ\)

Thus, we can say that, ABCD is a parallelogram and some angles are \(90^\circ\).

Hence it is proved that, ABCD is a rectangle.

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

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Given: A quadrilateral ABCD whose diagonals AC and BD bisect each other at right angles.

i.e., OA = OC and OB = OD

Also, \(\angle{AOD}\) = \(\angle{AOB}\) = \(\angle{COD}\) = \(\angle{BOC}\) = \(90^\circ\).

To prove: ABCD is a rhombus.

Proof: In \(\triangle{OAB}\) and \(\triangle{ODC}\), we have,

OA = OC and OB = OD ...(given)

\(\angle{AOB}\) = \(\angle{COD}\) ...(vertically opposite angles)

Therefore, \(\triangle{OAB}\) \(\displaystyle \cong \) \(\triangle{OCD}\) ...(By SAS rule)

Thus, AB = CD ...(i)(By CPCT)

Similarly, in \(\triangle{OAD}\) and \(\triangle{OBC}\), we have,

OA = OC and OD = OB ...(given)

\(\angle{AOD}\) = \(\angle{BOC}\) ...(vertically opposite angles)

Therefore, \(\triangle{OAD}\) \(\displaystyle \cong \) \(\triangle{OCB}\) ...(By SAS rule)

Thus, AD = BC ...(ii)(By CPCT)

Similarly, we can prove that,

AB = CD and

CD = BC ...(iii)

Hence, from (i), (ii) and (iii), we get,

AB = BC = AD = CD

Hence, it is proved that ABCD is a rhombus.

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

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Given: A square ABCD whose diagonals AC and BD intersect at O.

To prove: Diagonals are equal and bisect each other at right angles.

i.e, AC = BD, OD = OB, OA = OC and \({AC}\perp{BD}\)

Proof: In \(\triangle{ABC}\) and \(\triangle{BAD}\), we have,

BC = AD ...(given)

AB = BA ...(Common side)

\(\angle{ABC}\) = \(\angle{BAD}\) = \(90^\circ\)

Therefore, \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\) ...(By SAS rule)

Thus, AC = BD ...(By CPCT)

Similarly, in \(\triangle{OAB}\) and \(\triangle{OCD}\), we have,

AB = DC ...(given)

\(\angle{OAB}\) = \(\angle{DCO}\) ...(Since, AB || CD and transversal AC intersect)

\(\angle{OBA}\) = \(\angle{BDC}\) ...(Since, AB || CD and transversal BD intersect)

Therefore, \(\triangle{OAB}\) \(\displaystyle \cong \) \(\triangle{OCD}\) ...(By SAS rule)

Thus, OA = OC and OB = OD ...(By CPCT)

Now, in \(\triangle{AOB}\) and \(\triangle{AOD}\), we have,

OA = OD ...(proved earlier)

AB = AD ...(sides of square)

AO = OA ...(Commom side)

Therefore, \(\triangle{AOB}\) \(\displaystyle \cong \) \(\triangle{AOD}\) ...(By SSS rule)

Thus, \(\angle{AOB}\) = \(\angle{AOD}\) ...(By CPCT)

\(\angle{AOB}\) + \(\angle{AOD}\) = \(180^\circ\) ...(linear pair axiom)

\(\angle{AOB}\) = \(\angle{AOD}\) = \(90^\circ\)

Thus, \(AO\perp{BD}\), i.e., \(AC\perp{BD}\).

Also, AC = BD, OA = OC, OB = OD and \(AC\perp{BD}\)

Hence, it is proved that diagonals are equal and 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.

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Given: A quadrilateral ABCD in which AC = BD and AC \(\perp{BD}\) such that OA = OC
and OB = OD. So, ABCD is a parallelogram.

To prove: ABCD is a square.

Proof: Let AC and BD intersect at a point O.

In \(\triangle{ABO}\) and \(\triangle{ADO}\), we have,

BO = OD ...(given)

AO = OA ...(Common side)

\(\angle{AOB}\) = \(\angle{AOD}\) = \(90^\circ\) ...(given)

Therefore, \(\triangle{ABO}\) \(\displaystyle \cong \) \(\triangle{ADO}\) ...(By SAS rule)

Thus, AB = AD ...(By CPCT)

Also, AB = DC and AD = BC ...(opposite sides of parallelogram)

Therefore, AB = BC = DC = AD ...(i)

Similarly, in \(\triangle{ABC}\) and \(\triangle{BAD}\), we have,

AC = BD ...(given)

AB = BA ...(Common side)

BC = AD ...(from(i))

Therefore, \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\) ...(By SSS rule)

Thus, \(\angle{ABC}\) = \(\angle{BAD}\) ...(ii)(By CPCT)

But \(\angle{ABC}\) + \(\angle{BAD}\) = \(180^\circ\) ...(linear pair axiom)

\(\angle{ABC}\) = \(\angle{BAD}\) = \(90^\circ\) ...(from (ii))

Thus, AB = BC = CD = DA, and \(\angle{A}\) = \(90^\circ\)

Hence, it is proved that, ABCD is a square.

6. Diagonal AC of a parallelogram ABCD bisects \(\angle{A}\) (see figure). Show that

(i) It bisects \(\angle{C}\) also,

(ii) ABCD is a rhombus.

(i) It bisects \(\angle{C}\) also,

(ii) ABCD is a rhombus.

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Given, diagonal AC of a parallelogram ABCD bisects \(\angle{A}\).

i.e., \(\angle{DAC}\) = \(\angle{BAC}\) = 1/2 \(\angle{BAD}\) ...(i)

Here, AB || CD and AC is the transversal.

\(\angle{DCA}\) = \(\angle{CAB}\) ...(ii)(alternate angles)

\(\angle{BCA}\) = \(\angle{DAC}\) ...(iii)(alternate angles)

From eq. (i), (ii) and (iii), we get,

\(\angle{DAC}\) = \(\angle{BAC}\) = \(\angle{DCA}\) = \(\angle{BCA}\)

Now, \(\angle{BCD}\) = \(\angle{BCA}\) + \(\angle{DCA}\)

i.e., = \(\angle{DAC}\) + \(\angle{CAB}\)

Therefore, \(\angle{BCD}\) = \(\angle{BAD}\)

Thus, we can say that, diagonal AC bisects \(\angle{C}\).

Now, in \(\triangle{OAD}\) and \(\triangle{OCD}\), we have,

OA = OC ...(since, diagonals bisects each other)

DO = OD ...(Common side)

\(\angle{AOD}\) = \(\angle{COD}\) = \(90^\circ\)

Therefore, \(\triangle{OAD}\) \(\displaystyle \cong \) \(\triangle{OCD}\) ...(By SAS rule)

Thus, AD = CD ...(By CPCT)

Now, AB = CD and AD = BC ...(sides of parallelogram)

Therefore, AB = CD = AD = BC

Hence it is proved that, 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 \(\angle{D}\).

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Given: ABCD is a rhombus. i.e., AD = AB = BC = CD ... (i)

To prove: (i) Diagonal AC bisect \(\angle{A}\) as well as \(\angle{C}\).

(ii) Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

Proof: Let AC and BD are the diagonals of rhombus ABCD.

In \(\triangle{ABC}\) and \(\triangle{ADC}\), we have,

AD = AB ...(Given)

AC = CA ...(Common side)

CD = BC ...(From eq. (i))

Therefore, \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{ADC}\) ...(By SSS rule)

Thus, \(\angle{DAC}\) = \(\angle{BAC}\) ...(By CPCT)

Also, \(\angle{DCA}\) = \(\angle{BCA}\)

Also, \(\angle{DAC}\) = \(\angle{DCA}\)

And \(\angle{BAC}\) = \(\angle{BCA}\)

This shows that, Diagonal AC bisect \(\angle{A}\) as well as \(\angle{C}\).

Now, in \(\triangle{BDC}\) and \(\triangle{BDA}\), we have,

AB = BC ...(Given)

BD = BD ...(Common side)

AD = CD ...(Given)

Therefore, \(\triangle{BDC}\) \(\displaystyle \cong \) \(\triangle{BDA}\) ...(By SSS rule)

Thus, \(\angle{BDA}\) = \(\angle{BDC}\) ...(By CPCT)

Also, \(\angle{DBA}\) = \(\angle{DBC}\)

Also, \(\angle{BDA}\) = \(\angle{DBA}\)

And \(\angle{BDC}\) = \(\angle{DBC}\)

This shows that, Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

Hence, proved.

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 \(\angle{D}\).

(i) ABCD is a square

(ii) Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

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Given: ABCD is a rectangle. AB = CD and BC = AD ...(i)

To prove: (i) ABCD is a square. i.e., AB = BC = CD = DA

(ii) Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

Proof: In \(\triangle{ADC}\) and \(\triangle{ABC}\), we have,

\(\angle{DAC}\) = \(\angle{BAC}\) ...(Since, AB || DC and AC is transversal that intersects)

Similarly, \(\angle{DCA}\) = \(\angle{BCA}\)

AC = CA ...(Common side)

Therefore, \(\triangle{ADC}\) \(\displaystyle \cong \) \(\triangle{ABC}\) ...(By ASA rule)

Thus, AD = AB ...(By CPCT)

Also, CD = BC ...(ii)

Thus, from eq. (i) and(ii), we get,

AB = BC = AD = CD

Hence, it is proved that ABCD is a square.

Now, in \(\triangle{AOB}\) and \(\triangle{COB}\), we have,

AB = BC ...(Given)

BO = OB ...(Common side)

OA = OC ...(Since,diagonal bisectS each other)

Therefore, \(\triangle{AOB}\) \(\displaystyle \cong \) \(\triangle{COB}\) ...(By SSS rule)

Thus, \(\angle{OBA}\) = \(\angle{OBC}\) ...(By CPCT)

This shows that, Diagonal BD bisects \(\angle{B}\).

Similarly, Now, in \(\triangle{AOD}\) and \(\triangle{COD}\), we have,

AD = CD ...(Given)

OD = DO ...(Common side)

OA = OC ...(Since,diagonal bisectS each other)

Therefore, \(\triangle{AOD}\) \(\displaystyle \cong \) \(\triangle{COD}\) ...(By SSS rule)

Thus, \(\angle{ADO}\) = \(\angle{CDO}\) ...(By CPCT)

Hence, it is proved that, Diagonal BD bisects \(\angle{D}\), too.

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

i) \(\triangle{APD}\) \(\displaystyle \cong \) \(\triangle{CQB}\)

ii) AP = CQ

iii) \(\triangle{AQB}\) \(\displaystyle \cong \) \(\triangle{CPD}\)

iv) AQ = CP

v) APCQ is a parallelogram.

i) \(\triangle{APD}\) \(\displaystyle \cong \) \(\triangle{CQB}\)

ii) AP = CQ

iii) \(\triangle{AQB}\) \(\displaystyle \cong \) \(\triangle{CPD}\)

iv) AQ = CP

v) APCQ is a parallelogram.

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Given, ABCD is a parallelogram and P and Q are lie on BD such that DP = BQ ...(i)

i) Now,in \(\triangle{APD}\) and \(\triangle{CQB}\), we have,

DP = BQ ...(Given)

AD = BC ...(Opposite sides are equal in parallelogram)

\(\angle{ADP}\) = \(\angle{QBC}\) ...(Since, AD || BC and BD is a transversal)

Therefore, \(\triangle{APD}\) \(\displaystyle \cong \) \(\triangle{CQB}\) ...(By SAS rule)

ii)Since, \(\triangle{APD}\) \(\displaystyle \cong \) \(\triangle{CQB}\)

Therefore, AP = CQ ...(By CPCT)

iii) Here, Now, in \(\triangle{AQB}\) and \(\triangle{CPD}\), we have,

DP = BQ ...(Given)

AB = CD ...(Opposite sides are equal in parallelogram)

\(\angle{ABQ}\) = \(\angle{CDP}\) ...(Since, AB || CD and BD is a transversal)

Therefore, \(\triangle{AQB}\) \(\displaystyle \cong \) \(\triangle{CPD}\) ...(By SAS rule)

iv)Since, \(\triangle{AQB}\) \(\displaystyle \cong \) \(\triangle{CPD}\)

Therefore, AQ = CP...(By CPCT)

v) Now,in \(\triangle{APQ}\) and \(\triangle{PCQ}\), we have,

AQ = CP (From part iv))

AP = CQ (From part ii))

PQ = QP ...(Common side)

Therefore, \(\triangle{APQ}\) \(\displaystyle \cong \) \(\triangle{PCQ}\) ...(By SSS rule)

Thus, \(\angle{APQ}\) = \(\angle{PQC}\)

And \(\angle{AQP}\) = \(\angle{CPQ}\) ...(Vertically Opposite angles)

Now, these equal angles form a pair of alternate angle when line segment AP and QC are intersected by a transversal PQ.

i.e., AP || CQ and AQ || CP

Now, both pairs of opposite sides of quadrilateral APCQ are parallel.

Hence, it is proved that APCQ is a parallelogram.

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

(i) \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{CQD}\)

(ii) AP = CQ

(i) \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{CQD}\)

(ii) AP = CQ

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

Since, AB || CD and BD is a transversal, we get,

\(\angle{CDB}\) = \(\angle{DBA}\) ...(i)

Now, in \(\triangle{APB}\) and \(\triangle{CQD}\), we have,

CD = AB ...(Sides of parallelogram)

\(\angle{CQD}\) = \(\angle{APB}\) = \(90^\circ\) ...(Given)

\(\angle{CDQ}\) = \(\angle{ABP}\) ...(From Eq. (i))

Therefore, \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{CQD}\) ...(By ASA rule)

Thus, AP = CQ ...(By CPCT)

Hence, it is proved.

11. In \(\triangle{ABC}\) and \(\triangle{DEF}\), 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) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{DEF}\)

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) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{DEF}\)

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Given: In \(\triangle{ABC}\) and \(\triangle{DEF}\), AB = DE, AB || DE, BC = EF and BC || EF

(i) Now, in quadrilateral ABED,

AB = DE and AB || DE ...(Given)

Since, a pair of opposite sides is equal and parallel

Therefore, ABED is a parallelogram.

(ii) In quadrilateral BEFC,

BC = EF and BC || EF ...(Given)

Since, a pair of opposite sides is equal and parallel

Therefore, BEFC is a parallelogram.

(iii)Since, ABED is a parallelogram,

AD || BE and AD = BE ...(i)

Also, BEFC in a parallelogram,

CF || BE and CF = BE ...(ii)

Thus, from Eq. (i) and (ii), we get,

AD || CF and AD = CF ...(From part (iii))

Therefore, ACFD is a parallelogram.

(v)Since, ACFD is a parallelogram.

we get, AC = DF and AC || DF

(vi)Now, in \(\triangle{ABC}\) and \(\triangle{DEF}\),

AB = DE ...(Given)

BC = EF ...(Given)

and AC = DF ...(From part (v))

Therefore, \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{DEF}\) ...(By SSS test)

12. ABCD is a trapezium in which AB || CD and AD = BC (see figure).

Show that

i) \(\angle{A}\) = \(\angle{B}\)

ii) \(\angle{C}\) = \(\angle{D}\)

iii) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\)

iv) Diagonal AC = Diagonal BD

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

Show that

i) \(\angle{A}\) = \(\angle{B}\)

ii) \(\angle{C}\) = \(\angle{D}\)

iii) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\)

iv) Diagonal AC = Diagonal BD

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

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Given: ABCD is trapezium.

AB || CD and AD = BC

Now, extend AB and draw a line through C parallel to DA intersecting AB produced at E.

Now, ADCE is a parallelogram.

Therefore, AD || CE and AD = CE

But AD = BC

Therefore, AD = BC = CE

i) We know that, \(\angle{A}\) + \(\angle{E}\) = \(180^\circ\) ...(Since, interior angles on the same side of the transversal )

i.e., \(\angle{E}\) = \(180^\circ\) - \(\angle{A}\)

Therefore, \(\angle{E}\) = \(180^\circ\) - \(\angle{A}\) ...(Since, BC = EC)

Also, \(\angle{ABC}\) = \(180^\circ\) - \(\angle{CBE}\) ...(Since, ABE is straight line)

\(\angle{ABC}\) = \(180^\circ\) - \(180^\circ\) + \(\angle{A}\)

\(\angle{B}\) = \(\angle{A}\)

Hence, it is proved.

ii)Now, \(\angle{A}\) + \(\angle{D}\) = \(180^\circ\) ...(Since, interior angles on the same side of the transversal)

i.e., \(\angle{D}\) = \(180^\circ\) - \(\angle{A}\)

i.e., \(\angle{D}\) = \(180^\circ\) - \(\angle{B}\) ...(from eq.(i)) ...(ii)

Also, \(\angle{C}\) + \(\angle{B}\) = \(180^\circ\) ...180 (Since, interior angles on
the same side of the transversal BC)

i.e., \(\angle{C}\) = \(180^\circ\) - \(\angle{B}\) ...(iii)

from Eq. (ii) and (iii), we get,

\(\angle{C}\) = \(\angle{D}\)

Hence, proved.

iii)Now, in \(\triangle{ABC}\) and \(\triangle{BAD}\), we have,

AD = BC ...(Given)

\(\angle{A}\) = \(\angle{B}\) ...(From Eq.(i))

AB = BA ...(Common side)

Therefore, \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\) ...(By SAS rule)

Hence, proved.

iv)Since, \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\)

Therefore, AC = BD

Hence, it is proved.

1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC,
CD and DA (see figure). AC is a diagonal. Show that

i) SR || AC and SR = (1/2) AC

ii) PQ = SR

iii) PQRS is a parallelogram.

i) SR || AC and SR = (1/2) AC

ii) PQ = SR

iii) PQRS is a parallelogram.

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Given: P, Q, R, and S are mid-points of the sides.

Therefore, AP = PB, BQ = CQ
CR = DR and AS = DS

i) Now, in \(\triangle{ADC}\), we have,

S is the midpoint of AD and R is the mid point of CD.

As, we know that,

By midpoint theorem, the line segment joining the mid points of two sides of a triangle is parallel to the third side.

Thus, we can say that, SR || AC ...(i)

and SR = (1/2) AC ...(ii)

ii) Similarly, now, in \(\triangle{ABC}\), we have,

PQ || AC ...(iii)

and PQ = (1/2) AC ...(iv)

Now, from (ii) and (iv), we get,

SR = PQ = (1/2) AC ...(v)

Now, from (i), (iii) and (v), we get,

PQ || SR and PQ = SR.

Therefore, PQRS is a parallelogram ...(Since, a pair of opposite sides of a quadrilateral PQRS is equal and parallel)

Hence, proved.

2. 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.

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Given: ABCD is a rhombus and P, Q, R and S are mid-points of AB, BC, CD and DA.

By midpoint theorm, Now, in \(\triangle{ADC}\), we have,

Thus, SR || AC ...(i)

and SR = (1/2) AC ...(ii)

Similarly, in \(\triangle{ABC}\), we have,

PQ || AC ...(iii)

and PQ = (1/2) AC ...(iv)

from (i), (ii), (iii) and (iv), we get,

SR = PQ = (1/2) AC and PQ || SR

Therefore, Quadrilateral PQRS is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at right angles.

Therefore, \(\angle{EOF}\) = \(90^\circ\).

Now, By midpoint theorem, we have,

RQ || BD

Thus, RE || OF

As SR || AC ...(from (i))

Thus, FR || OE

Therefore, OERF is a parallelogram

So, \(\angle{ERF}\) = \(\angle{EOF}\) = \(90^\circ\) ...(since, Opposite angle of a quadrilateral is equal)

Thus, PQRS is a parallelogram with \(\angle{R}\) = \(90^\circ\)

Hence, it is proved that PQRS is a rectangle.

3. 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.

Show that the quadrilateral PQRS is a rhombus.

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Given: ABCD is a rectangle.

i.e., \(\angle{A}\) = \(\angle{B}\) = \(\angle{C}\) = \(\angle{D}\) = \(90^\circ\) and AB = CD and BC = AD.

Also, P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.

Therefore, by midpoint theorem,

PQ || BD and PQ = (1/2) BD

and SR || AC and SR = (1/2) AC

In rectangle ABCD,

AC = BD

Thus, PQ = SR ...(i)

Now, in \(\triangle{ASP}\) and \(\triangle{BQP}\), we have,

AP = BP ...(Given)

AS = BQ ...(Given)

\(\angle{A}\) = \(\angle{B}\) ...(Given)

Therefore, \(\triangle{ASP}\) \(\displaystyle \cong \) \(\triangle{BQP}\) ...(By SAS rule)

Therefore, SP = BQ ...(ii)(By CPCT)

Similarly, in \(\triangle{RDS}\) and \(\triangle{RCQ}\), we have,

SD = CQ ...(Given)

DR = RC ...(Given)

\(\angle{C}\) = \(\angle{D}\) ...(Given)

Therefore, \(\triangle{RDS}\) \(\displaystyle \cong \) \(\triangle{RCQ}\) ...(By SAS rule)

Therefore, SR = RQ ...(iii)(By CPCT)

Thus, from eq. (i), (ii) and (iii), we can say that,

The quadrilateral PQRS is a rhombus is proved.

4. 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.

A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC.

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Given: ABCD is a trapezium in which AB || CD and E is mid-point of AD and EF || AB.

In \(\triangle{ABD}\), we have,

EP || AB and E is mid-point of AD.

So, by theorem, if a line drawn through the
mid-point of one side of a triangle is parallel to another side, then it bisects the third side.

Therefore, P is the midpoint of BD.

Similarly, in \(\triangle{BCD}\),

we have, PF || CD,

Therefore by converse of mid point theorem, F is mid-point of BD.

5. 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.

Show that the line segments AF and EC trisect the diagonal BD.

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

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

Proof: Since, ABCD is a parallelogram,

AB || DC and

AB = DC ...(since, Opposite sides of a parallelogram)

Thus, AE || FC and (1/2) AB = (1/2) CD

i.e., AE = FC

Therefore, AECF is a parallelogram.

Thus, AF || EC

And hence, EQ || AP and FP || CQ.

In \(\triangle{BAP}\), E is the mid-point of AB and EQ || AP,

So, By converse of mid-point theorem,

Q is the mid-point of BP.

i.e., BQ = PQ ...(i)

Similarly, in \(\triangle{DQC}\), F is the mid-point of DC and FP || CQ,

So, P is the mid-point of DQ.

i.e., DP = PQ ...(ii)

From Equations (i) and (ii) , we get,

BQ = PQ = DP

Hence, it is proved that CE and AF trisect the diagonal BD.

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

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Let ABCD is a quadrilateral and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively.

i.e., AS = SD, AP = BP, BQ = CQ and CR = DR.

We have to show that: PR and SQ bisect each other i.e., SO = OQ and PO = OR.

Now, in \(\triangle{ADC}\), S and R are mid-point of AD and CD.

We know that, the line segment joining the mid-points of two sides of a triangle is
parallel to the third side.

Thus, by midpoint theorem,

SR || AC and SR = (1/2) AC ...(i)

Similarly, in \(\triangle{ABC}\), P and Q are mid-point of AB and BC.

Thus, by midpoint theorem,

PQ || AC and PQ = (1/2) AC ...(ii)

From Eq. (i) and (ii), we get,

i.e., PQ || SR and PQ = SR = (1/2) AC

Therefore, Quadrilateral PQRS is a parallelogram whose diagonals SQ are PR.

Also, we know that diagonals of a parallelogram bisect each other. So, and bisect each other.

Hence, it is proved that PR and SQ bisect each other.

7. 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 is perpendicular to AC

iii) CM = MA = (1/2) AB

Show that

i) D is the mid-point of AC

ii) MD is perpendicular to AC

iii) CM = MA = (1/2) AB

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Given: \(\triangle{ABC}\) is a right angled triangle i.e., \(\angle{C}\) = \(90^\circ\)

and M is the mid-point of AB.

Also, DM || BC

i) In \(\triangle{ABC}\), M is the mid-point of AB and BC || MD,

So, By converse of mid-point theorem,

D is the mid-point of AC.

i.e., AD = CD ...(i)

ii) Since, BC || MD and CD is transversal.

Therefore, \(\angle{ADM}\) = \(\angle{ACB}\) ...(Corresponding angles)

But, \(\angle{C}\) = \(90^\circ\)

Therefore, \(\angle{ADM}\) = \(90^\circ\)

Hence, proved that MD is perpendicular to AC.

iii) Now, in \(\triangle{ADM}\) and \(\triangle{ACM}\), we have,

AD = CD ...(from (i))

DM = MD ...(Common side)

\(\angle{ADM}\) = \(\angle{MDC}\) = \(90^\circ\) ...(Proved)

Therefore, \(\triangle{ADM}\) \(\displaystyle \cong \) \(\triangle{CDM}\) ...(By SAS rule)

Therefore, CM = AM ...(ii)(By CPCT)

Also, M is the mid point of AB.

i.e., AM = BM = (1/2) AB ...(iii)

Thus, from (ii) and (iii),

CM = AM = (1/2) AB

Hence, proved.