D and E are the mid-points of AB and BC respectively.
ஃ DE II AC
Similarly, DF II BC and EF II AB
ஃ ADEF, BDFE, and DFCE are all parallelograms
DE is the diagonal of parallelogram BDFE
ஃ ΔBDE ≡ ΔFED
ΔDAF ≡ ΔFED
ΔEFC ≡ ΔFED
ஃ All the four triangles are congruent.

In ΔADG, E is the mid-point of AD and EF || DG.
⇒ F is the mid-point of AC (converse of mid-point theorem)
⇒ AF=FG...(i)
In ΔCBF, BF II DG, D is the mid-point of BC
⇒ G is the mid-point of FC
FG = GC...(ii)
From (i) and (ii), AF = FG = GC

AC = AF + FG + GC
= 3AF.

Let ABCD is a parallelogram
To show LMNO is a rectangle,
ㄥA+ㄥD=180°
\(\frac{1}{2}\)ㄥA+\(\frac{1}{2}\)ㄥD=90°
ㄥOAD+ㄥODA=90°
In ΔOAD,
ㄥOAD+ㄥADO+ㄥDOA=180°
⇒ ㄥDOA = 90°
⇒ ㄥLON = 90°
Similarly, ㄥOLM =ㄥLMN =ㄥMNO = 90°

ஃ A quadrilateral with all angles 90° is a rectangle. Also opposite angles are equal. It is rectangle.

Let ABCD is a parallelogram
To show LMNO is a rectangle,
ㄥA+ㄥD=180°
\(\frac{1}{2}\)ㄥA+\(\frac{1}{2}\)ㄥD=90°
ㄥOAD+ㄥODA=90°
In ΔOAD,
ㄥOAD+ㄥADO+ㄥDOA=180°
⇒ ㄥDOA=90°
⇒ ㄥLON=90°
Similarly, ㄥOLM=ㄥLMN=ㄥMNO=90°

ஃ A quadrilateral with all angles, 90° is a rectangle. Also, opposite angles are equal. It is a rectangle.

ㄥ1=ㄥ2=\(\frac{1}{2}\)ㄥA
ㄥ3=ㄥ4=\(\frac{1}{2}\)ㄥC
In ΔCXB,
ㄥ3+ㄥX+ㄥB=180° ...(i)
(By Angle sum property of a Δ)
In ΔDAY,
ㄥ1+ㄥY+ㄥD=180° ...(ii)
(By Angle sum property of a Δ)
Adding (i) and (ii),
ㄥ3+ㄥX+ㄥB+ㄥ1+ㄥY+ㄥD=180°+180°
i.e., ㄥX+ㄥY+ㄥ3+ㄥ1+ㄥB+ㄥD=180
i.e., ㄥX+ㄥY+\(\frac{1}{2}\)ㄥC+\(\frac{1}{2}\)ㄥA+ㄥB+ㄥD=360...(iii)
But, ㄥA+ㄥB+ㄥC+ㄥD=360...(iv)
(Angle sum property of a quadrilateral)
From (iii) and (iv),
∠X+ㄥY+\(\frac{1}{2}\)ㄥC+\(\frac{1}{2}\)ㄥA+ㄥ​​​​B+ㄥD=ㄥA+ㄥB+ㄥC+ㄥD
i.e., ㄥX+ㄥY=ㄥA-\(\frac{1}{2}\)ㄥA+ㄥC-\(\frac{1}{2}\)ㄥC
=\(\frac{1}{2}\)ㄥA+\(\frac{1}{2}\)ㄥC
ㄥX+ㄥY=\(\frac{1}{2}\)(ㄥA+ㄥC)
Hence Proved.

(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 triangle, ABC, points M and N on the sides AB and AC respectively are taken so that AM
= \(1\over4\) AB and AN = \(1\over4\)AC
To prove:
MN = \(1\over4\) BC.

Construction: Join EF where E and F are the mid-points of AB and AC respectively.
Proof:  \(\therefore\) E is the mid-point of AB and F is the mid-point of AC.
\(\because\)  EF IIBC and EF =  \(1\over2\)BC    ...........(1)
Now  AE = \(1\over2\) AB and AM = \(1\over4\) AB AM = \(1\over2\)AE
Similarly,    AN = \(1\over2\)AF
 ⇒ M and N are the mid-points of AE and AF respectively.
\(\therefore\) MN II EFand MN= \(1\over2\)EF= \(1\over2\) (\(1\over2\)BC)  I From (1)
 ⇒ = \(1\over4\)BC

Given:    ABCD is a parallelogram. AC is a diagonal of parallelogram ABCD which divides it into two triangles, namely, \(\Delta \) ABC and \(\Delta \)CDA

To Prove: \(\Delta \)ABC  \(\cong \) \(\Delta \)CDA
Proof: BC IIDA  IOpposite sides of a parallelogram are parallel and    AC   is a transversal
 \(\angle \)  BCA=\(\angle \)DAC  ...(1)  I Pair of alternate interior angles  
 Also, AB IIDC     IOpposite sides of a parallelogram are parallel and AC is a transversal
\(\therefore \) \(​​\angle \)BAC = \(\angle \)DCA   ......(2)   I Alternate interior angles      
AC = CA      ...(3)    I Common  
In view of (1), (2) and (3),
\(\Delta \) ABC \(\cong \) \(\Delta \) CDA      I ASA congruence criterion

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.