We draw the figure as shown below.

Let P, Q, R, and S be the mid-points of the sides AB, BC, CD, and DA respectively. Join AC and BD.Which intersects each other at O.
Now, in \(\Delta ABC, PQ\parallel AC\ and\ PQ=\frac { 1 }{ 2 } AC \ ...........(i)\)
[ by mid-point theorem ]
and in \(\Delta ACD,\ RS\parallel AC\ and \ RS=\frac { 1 }{ 2 } AC \ .........(ii)\)
From Eqs. (i) and (ii) we get,  [by mid-point theorem]
\(PQ\parallel RS\ and\ PQ\ =\ RS\)
[since a pair of opposite sides equal and parallel]
So, PQRS is a parallelogram
Also, AB = AD  [given] ...(i)
So, A lies on the perpendicular bisector of BD further
CB = CD [given]....(ii)
So, C lies on the perpendicular bisector of BD......(iii)
From Eqs. (ii) and (iii), AC is the perpendicular bisector of BD.
i.e.  AC\(\bot \)BD  \(\Rightarrow \angle AOD=90°\)
Now, clearly HS\(\parallel \)OE and SE\(\parallel \) OH
So, SEOH is a parallelogram.
Hence, \(\angle ESH=\angle EOH=90°\)
So, parallelogram PQRS is a rectangle.
Hence proved.

Given, PQRS is a rhombus \(\)\(\Rightarrow\) PQRS is a parallelogram
Since, PQ\(\)\(\parallel \) SR and PR is a transversal.
\(\therefore\) \(\angle ORS=\angle OPQ=35°\)  [alternate interior angles]....(i)
Also, the diagonals of a rhombus bisect each other at a right angle.
\(\therefore \angle SOR=90°\)                                                      .....(ii)
Now, in \(\Delta SOR,\angle RSO+\angle SOR+\angle ORS=180°\)
[angle sum property of triangles]
\(\Rightarrow\angle RSO=180°-\angle SOR-\angle ORS\)
\(\Rightarrow \angle RSO=180°-90°-35°\) [using Eqs. (i) and (ii)]
\(\Rightarrow \angle RSO=55°\)
\(\Rightarrow \angle OQP=55°\)
[alternate interior angles, Since PQ\(\parallel \) SR and QS is transversal]
From Eqs. (i) and (iii), we get
\(\angle ORS+\angle OQP=35°+55°=90°\) 

Lines l and m are parallel, i.e. PS\(\parallel \)QR and transversal t intersects PS and QR at points A and C, respectively. The bisectors of \(\angle\)PAC and \(\angle\)ACQ intersect each other at B and bisectors of \(\angle\)ACR and \(\angle\)SAC intersect each other at D. We have to show that, quadrilateral ABCD is a rectangle.
Since PS\(\parallel \)CR and t is transversal.
\(\therefore\)  \(\angle\)PAC =\(\angle\)ACR  [alternate angles]
\(\Rightarrow \frac { 1 }{ 2 } \angle PAC=\frac { 1 }{ 2 } \angle ACR\)
\(\Rightarrow \angle BAC=\angle ACD\)
These form a pair of alternate angles for lines AB and CD with AC as transversal.
So,  AB\(\parallel \)DC
Similarly, BC\(\parallel \) AD
Therefore, quadrilateral ABCD is a parallelogram.
Now, PS is a straight line.
So,  \(\angle\)PAC + \(\angle\)SAC = 180\(°\)
[linear pair axiom]
\(\Rightarrow \frac { 1 }{ 2 } \angle PAC+\frac { 1 }{ 2 } \angle SAC=\frac { 1 }{ 2 } \times 180°\)
\(\left[ multiply\ by\frac { 1 }{ 2 } on\ both\ sides \right] \)
\(\Rightarrow \angle BAC+\angle DAC=90°\)
\(\Rightarrow \angle BAD=90°\)
Thus, one angle of parallelogram ABCD is a right angle.
Hence, ABCD is a rectangle.                          
Hence Proved.

In quadrilateral ABCD,
\(\angle A+\angle C=140° \ .....(i)\)
\(\angle A:\angle C=1:3\quad \ ....(ii)\)
and  \(\angle B:\angle D=5:6\ ....(iii)\)
Let \(\angle\)A=x and \(\angle\)C = 3x
Then, from Eq. (i), we get
x + 3x = 140\(°\)
\(\Rightarrow\) 4x = 140\(°\)
\(\Rightarrow\) x = \(\frac { 140° }{ 4 } =35°\)
\(\therefore\) \(\angle\)A = 35 \(°\)
and \(\angle\)C = 3 \(\times \)35\(°\) =105\(°\)
We know that, sum of angles of a quadrilateral = 360\(°\)
\(\therefore\) \(\angle\)A + \(\angle\)B + \(\angle\)C + \(\angle\)D = 360\(°\)
\(\Rightarrow\)\(\angle\)B + \(\angle\)D = 360\(°\) -( \(\angle\)A + \(\angle\)C)
\(\Rightarrow\)\(\angle\)B +  \(\angle\)D = 360\(°\) - (35\(°\) + 105\(°\)) = 220\(°\)
Since, \(\angle\)B : \(\angle\)D = 5: 6
\(\therefore\angle B=\frac { 5 }{ 5+6 } \times 220°=\frac { 5 }{ 11 } \times 220°=100°\)
\(and\ \angle D=\frac { 6 }{ 5+6 } \times 220°\)
\(=\frac { 6 }{ 11 } \times 220°\)
\(=120°\) 
Hence, \(\angle A=35°,\angle B=100°,\angle C=105°\quad and\quad \angle D=120°\) 

Given: The diagonals AC and BD of a quadrilateral ABCD are equal and bisect each other at right angles.
To Prove: Quadrilateral ABCD is a square.
Proof: In \(\Delta\)OAD and \(\Delta\)OCB,
OA = OC             I Given
OD = OB       I Given
\(\angle AOD=\angle COB\)    |Vertically Opposite Angles
 \(\therefore \Delta OAD\cong \Delta OCB\)    I SAS Congruence Rule

\(\therefore \)   AD = CB           I C.P.C.T.
 \(\angle ODA=\angle OBC\)  |C.P.C.T
\(\angle BDA=\angle DBC\)
.\(\therefore \)   AD II BC
Now, \(\because\) AD = CB and AD II CB
\(\therefore \)  Quadrilateral ABCD is a II gm.      I A quadrilateral is a parallelogram if a pair of opposite sides are parallel and  equal.
In \(\therefore \) \(\Delta\) AOB and \(\Delta\)AOD,
AO = AO                          I Common
OB = OD          I Given
 \(\angle \)AOB = \(\angle \)AOD               |Each=\(90°\)
\(\therefore \)  \(\Delta\)AOB \(\cong \)  \(\Delta\) AOD         I SAS Congruence Rule
AB = AD                 I C.P.C.T.
Now, \(\because\) ABCD is a parallelogram and
AB=AD
 \(\because\) ABCD is a rhombus.
Again, in \(\Delta\) ABC and \(\Delta\) BAD,
AC = BD            I Given
BC = AD            I \(\because\)  ABCD is a rhombus
AB =  BA                     I Common
\(\therefore \) \(\Delta\) ABC \(\cong \) \(\Delta\) BAD       I SSS Congruence Rule
\(\therefore \)    \(\angle \) ABC = \(\angle\) BAD    I C.P.C.T.
\(\because\)    AD II BC                 I Opp. sides of IIgm ABCD and transversal AB intersects them.
\(\therefore \) \(\angle ABC+\angle BAD=180°\)      I Sum of consecutive interior angles on the  same side of a transversal is 180°
ABC = \(\angle\)BAD = \(90°\)
Similarly,\(\angle\)BCD =\(\angle\)ADC =\(90°\)
 \(\therefore \)  ABCD is a square.

Given ABCD is a rectangle.
\(\therefore\) AB = DC 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
(i) In \(\Delta \)CDA and \(\Delta \)ABC,
\(\angle\)DAC = \(\angle\)BCA
[Since, BC II AD andACis a transversal]
\(\angle\)DCA = \(\angle\)BAC
[Since, AB II DC and AC is a transversal]
and AC = CA [common side]
\(\therefore\)\(\Delta \)CDA \(\cong \) \(\Delta \)ABC [by ASA congruence rule]
Then, AD = AB [by CPCT]
and CD = BC ... (ii)
From Eqs. (i) and (ii), we get
AB = BC = AD = CD
So, ABCD is a square.
(ii) In \(\Delta \)AOB and \(\Delta \)COB, we have
AB=BC [sides of a square]
BO=OB   [common side]
OA=OC
[Since, diagonals of a square bisect each other]
\(\therefore\) \(\Delta \)AOB \(\cong \) \(\Delta \)COB [by SSS congruence rule]
Then,\(\angle\)OBA = \(\angle\)OBC [byCPCT]
This shows that BO or BD bisects \(\angle\)B.
Similarly, in \(\Delta \)AOD and \(\Delta \)COD, we have
AD = CD [sides of a square]
OD = DO [common side]
and OA = OC
[Since, diagonals of square bisect each other]
\(\therefore\) \(\Delta \)AOD \(\cong \) \(\Delta \)COD [by SSS congruence rule]
Then, \(\angle\)ADO = \(\angle\)CDO [by CPCT]
This shows that DO or DB bisect \(\angle\)D.
Hence, BD bisects \(\angle\)B as well as \(\angle\)D. Hence proved.

Given: ABC is an isosceles triangle in which AB = AC. AD bisects L PAC and CD IIAB.
To Prove:
(i)  \(\angle \)DAC =\(\angle \)BCA
(ii) ABCD is a parallelogram.
Proof:
(i) In \(\Delta \) ABC,
\(\because\) AB = AC
\(\therefore\) \(\angle \)B =\(\angle \)C               .......(1)      I Angles opposite to equal sides of a triangle are equal
Also, Ext.   \(\angle \)PAC =\(\angle \)B +\(\angle \)C     I An exterior angle of a triangle is equal to the sum of its two interior opposite angles
⇒    \(\angle \)PAC = \(\angle \)C+\(\angle \)C   | From (1)
⇒ 2\(\angle \)CAD = 2\(\angle \)C   | \(\because\) AD bisects \(\angle \)PAC 
⇒ \(\angle \)CAD =\(\angle \)C
⇒  \(\angle \) DAC =\(\angle \)BCA
(ii) But these angles form a pair of equal alternate interior angles
\(\therefore\)  AD II BC
Also, CD II AB         I Given
\(\therefore\)   ABCD is a parallelogram   IA quadrilateral is a parallelogram if its both the pairs of opposite sides are parallel.

AB = 10 cm

Sum of the angles of a quadrilateral is 360°.
ஃ 4x° + 7x° + 15x° + 10x° = 360°
[Angle sum property of quadrilateral]
⇒ 36x°= 360°
⇒ x=10°
ஃ Smallest angle = 4x° = 40°
Largest angle = 15x°= 150°

\(\angle A={ 60 }^{ ° },\angle B={ 120 }^{ ° },\angle C={ 60 }^{ ° },\angle D={ 120 }^{ ° }\)

Given ABCD is a square and \(\angle COD={ 105 }^{ ° }\)
We know that, the diagonal AC of square ABCD will bisect the \(\angle C\).
\(\therefore\angle OCX=\frac { 1 }{ 2 } \angle C=\frac { 1 }{ 2 } \times { 90 }^{ ° }={ 45 }^{ ° }\)........... (i)
Now, we have
\(\angle COD+\angle COX={ 180 }^{ ° }\)[linear pair axioms]
\(\Rightarrow { 105 }^{ ° }+\angle COX={ 180 }^{ ° }\)
\(\Rightarrow \angle COX={ 180 }^{ ° }-{ 105 }^{ ° }\)
\(\Rightarrow \quad \angle COX={ 75 }^{ ° }\)           .... (ii)
Now, in \(\triangle COX\),
\(\angle COX+\angle OXC+\angle OCX={ 180 }^{ ° }\)
[angle sum property of a triangle]
\(\Rightarrow { 75 }^{ ° }+x+{ 45 }^{ ° }={ 180 }^{ ° }\) [using Eqs.(i) and (ii)]
\(\Rightarrow { x= }{ 180 }^{ ° }-{ 120 }^{ ° }\)
\(\therefore{ x= }{ 60 }^{ ° }\)

Given ABCD is a parallelogram, whose diagonals bisect each other at O, i.e. OB = OD and OC = OA.
To prove PQ is bisected at O.
Proof In \(\triangle ODP\ and\ triangle \ OBQ\) ,
\(\angle POD=\angle BOQ\)
[vertically opposite angles]
\(\angle ODP=\angle OBQ\)
[since, \(AD\parallel BC\) and BD is transversal]
and OD = OB
\(\therefore \triangle OBQ\cong \triangle ODP\)   [by ASA congruence rule]
Then, OQ = OP       [by CPCT]
Hence, O bisects PQ.

10.5 cm

Theorm

In a trapezium, only one pair of opposite sides is parallel.

We are given that AB = BC and have to prove that
DE = EF.
Let us join A to E intersecting m at G.
Let trapezium ACFD is divided into two triangles, namely ΔACF and ΔAFD.
In ΔACF, it is given that B is the mid-point of AC(AB = BC) and BG II CF (Since m || n)
So, G is the mid-point of AF (By the converse of midpoint theorem)
Now in ΔAFD, we can apply the sam argument as G is the mid-point of AF, GE IIAD so E is the mid-point of DF
i.e., DE = EF
In otherwords l, m and n cut off equal intercepts on q also.

(i) As EB II DL and ED II BL.
Therefore EBLD is a parallelogram.
ஃ BL = ED
=\(\frac{1}{2}\)BC=CL ...(i)
Now in triangles DCL and FBL, we have
CL=BL from (i)
ㄥDLC= ㄥFLB 
ㄥCDL=ㄥBFL
ΔCDL≡ΔBFL
CD= BF
and DL= FL
Now, BF = DC = AB
⇒ 2AB = 2DC
⇒ AF = 2DC
(ii) ∵ DL=FL
⇒ DF = 2DL