55°

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 \(\Delta\)ABC,
\(\because\) R and Q are the mid-points of AB and BC respectively.
\(\therefore\) RQ II AC and RQ = \(1\over2\)AC.
Similarly, we can show that
PS II AC and PS = \(1\over2\) AC
\(\therefore\) RQ II PS and RQ = PS.
Thus a pair of opposite sides of a quadrilateral PQRS are parallel and equal.
\(\therefore\)  PQRS is a parallelogram.
Since the diagonals of a parallelogram bisect each other.
\(\therefore\) PR and QS bisect each other.

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively.
To Prove:
\((i)\Delta APB\cong \Delta CQD\)
\(\\ (ii)AP=CQ\)
Proof: (i) In \(\Delta APB\ \) and \( \Delta CQD\)
AB = CD                I Opp. sides of || gm ABCD
\(\angle ABP=\angle CDQ\)           |AB || DC and transversal BD intersects them
 \(\angle ABP=\angle CDQ\)           |Each=\(90°\)
(ii)  \(\Delta APB\cong \Delta CQD\)                I Proved above in (i)
 \(\therefore \)  AP = CQ.                I C.P.C.T.

Given: Diagonal AC of a parallelogram ABCD bisects \(\angle A\)
To Prove:
(i) It bisects \(\angle C\) also
(ii) ABCD is a rhombus.
Proof: (i) In \(\Delta\)ADC and \(\Delta\)CBA,
AD = CB                 I Opp. sides of IIgm ABCD
CA = AC               I Common
DC = BA               I Opp. sides of II gm ABCD
\(\therefore \)  \(\Delta\)ADC \(\cong \) \(\Delta\)CBA               I SSS Congruence Rule
\(\angle ACD=\angle CAB\)     |C.P.C.T
 \(\angle DAC=\angle BCA\)     |C.P.C.T
but  \(\angle CAB=\angle DAC\)     |Given
 \(\therefore \ \angle ACD=\angle BCA\)
 \(\therefore \) AC bisects \(\angle C\) also
(ii) From above,
\(\angle ACD=\angle CAD\)  
AD = CD            I Sides opposite to equal angles  of a triangle are equal
\(\therefore \) AB = BC = CD = DA         I \(\therefore \) ABCD is a II gm
\(\therefore \)  ABCD is a rhombus.

Given: ABCD is a parallelogram and E is the mid-point of side BC. DE and AB on producing meet at F.
To Prove: AF = 2AB
Proof: In  \(\Delta \)FAD,
\(\therefore\) E is the mid-point of BC        I Given
and EB II DA   | Opposite sides of a parallelogram are parallel
\(\therefore\) B is the mid-point of AF     I By converse of mid-point theorem
AB = BF = \(1\over2\) AF ⇒ AF = 2AB

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.

10.5 cm

5 cm

72°, 108°, 72°, 108°

\(36°\),\(144°\),\(36°\), \(144°\)

40°, 80°, 100°, 140°; Trapezium

FE IIBC, ED II AB
\(\therefore\)BDEF is a parallelogram

\(\because \)   ABD=\(90°\),\(\therefore \)   FED = \(90°\)

Perimeter of quadrilateral BDEF
= BD + DE+EF+FD
= 2 (BD + DE)
= 2 (\(1\over2\)BC + \(1\over2\)AB)
= BC + AB = 4 + 3 = 7 cm.

\(\angle \)GAE = \(\angle \)DCB =\(60°\)
\(\therefore \) \(\angle \)GFE = \(\angle \)GAE =\(60°\)

See a rhombus

no of angle of a rhombus is 90⁰

opposite sides are not equal in a kite