In Δs RSM and PQL, 

ㄥRSM=ㄥPQL
ㄥM=ㄥL=90°
SR=PQ
By AAS, ΔRSM≡ΔPQL
(ii) PL=RM(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.

(i) Since ABCD is a square and ΔDCE is an equilateral triangle.

ㄥADC=90°
and ㄥEDC=60°
ㄥADC+ㄥEDC=90°+60°
⇒ ㄥADE=150°
Similarly, we have
ㄥBCE=150°
Thus in ∆ADE and ΔBCE, we have
AD= BC
ㄥADE=ㄥBCE=150°
and DE = CE
So by SAS congruence criterion, we have
Δ ADE≡Δ BCE
⇒ AE = BE Proved.
(ii) In ΔEAD, we have
AD= DE
⇒ ㄥEAD=ㄥAED=x(say)
Now, ㄥADE+ㄥAED+ㄥDAE=180°
150° + x + x = 180°
2x = 180°- 150° = 30°
x=ㄥDAE=15° Proved.

Given: In a quadrilateral ABCD, the line segments bisecting \(\angle \)C and \(\angle \)D meet at E.
  To Prove:   \(​​\angle \)A+\(​​\angle \)B = 2 \(\angle \)CED
I Angle sum property of a quadrilateral
In \(\Delta \) CED,
\(\angle \)CED +\(\angle \)EDC+\(\angle \)ECD = \(180°\)   | Angle sum property of a triangle
⇒ \(\angle \)CED+\(1\over2\)\(\angle \)D+\(1\over2\)\(\angle \)C=\(180°\)
 ⇒ 2\(\angle \)CED +\(\angle \) D+\(\angle \)C =\(360°\)    ...(2)
From (1) .and (2),
2\(\angle \)CED+\(\angle \)D+C=A+B+C+D
 ⇒ 2 \(\angle \)CED =\(\angle \)A+\(\angle \)B
 ⇒ \(\angle \)A+\(\angle \)B=2\(\angle \)CED

Let us recall what a rectangle is.
A rectangle is a parallelogram in which one angle is a right angle.
 
Let ABCD be a rectangle in which \(\angle\) A = 90°.
We have to show that \(\angle\) B = Ð C = \(\angle\) D = 90°
We have, AD || BC and AB is a transversal
(see Fig.).
So, \(\angle\) A + \(\angle\) B = 180° (Interior angles on the same
side of the transversal)
But, \(\angle\) A = 90°
So, \(\angle\) B = 180° – \(\angle\) A = 180° – 90° = 90°
Now, \(\angle\) C = Ð A and \(\angle\) D = \(\angle\) B
(Opposite angles of the parallellogram)
So, \(\angle\) C = 90° and \(\angle\) D = 90°.
Therefore, each of the angles of a rectangle is a right angle.

AM丄DC, AN丄BC
In quadrilateral AMCN,
ㄥA+ㄥM+ㄥC+ㄥN=360°
ㄥA+ㄥC=180°
⇒ 50°+ㄥC=180° ⇒ ㄥC=130°

In parallelogram, ㄥA=ㄥC=130°
ㄥB=ㄥD=180°-130°
=50°

Let the measure of the angles be x, 2.x, 3x and 4x then,
x + 2x + 3x + 4x = 360°
⇒ x = 36°
ஃ Angles of quadrilateral are 36°, 72°, 108°,144°

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°

72°, 108°, 72°, 108°

Theorem

\(BC=\sqrt { { BE }^{ 2 }+{ CE }^{ 2 } } =\sqrt { { \left( \frac { BD }{ 2 } \right) }^{ 2 }+{ \left( \frac { CA }{ 2 } \right) }^{ 2 } } \)\(=\frac { 1 }{ 2 } \sqrt { { BD }^{ 2 }+{ CA }^{ 2 } } =\frac { 1 }{ 2 } \sqrt { { 6 }^{ 2 }+{ 8 }^{ 2 } } =5\quad c.m\)

x+\(80°\)=3x- \(1 0°\)\(\Rightarrow \)   x=\(45°\)

no of angle of a rhombus is 90⁰

(i) Let the photo-frame be ABC such that BC = a, CA = band AB = c and the mid-points of AB, BC and CA are respectively D, E and F.
We have to determine the perimeter of ΔDEF
In ΔABC, DF is the line-segment joining the mid-points of sides AB and AC.

So, DF is parallel to BC and half of it.
i.e., DF=\(\frac{BC}{2}\)=\(\frac{a}{2}\)
Similarly, DE=\(\frac{AC}{2}=\frac{b}{2}\)
and EF=\(\frac{AB}{2}=\frac{c}{2}\)
DF + DE + EF=\(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}\)=\(\frac{a+b+c}{2}\)
Hence required perimeter=\(\frac{1}{2}\)(a+b+c)
(ii) Mid-point theorem.
(iii) Unity and cooperation or Mutual understanding.

(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