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.

According to the question, E and F are the midpoints of sides AB and CD.
ஃ AE=\(\frac{1}{2}\)AB
CF=\(\frac{1}{2}\)CD
ஃ In the parallelogram opposite sides are equal, so AB=CD
ஃ AE=CF
Again, AB||CD
So, AE||FC
Hence AECF is a parallelogram.
In ΔABP,
E is the mid-point of AB.EQ || AP
ஃ Q is the mid-point of BP
Similarly, P is the mid-point of DQ
DP= PQ= QB
ஃ Line segments AF and EC trisect the diagonal BD.

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.

In trapezium ABCD,
AB II DC and AD = BC
Through C, draw
CE II DA.
DC II AE and CE is transversal.

∠1=∠2
∠3=∠1
∠2=∠3=∠1
∠2+∠3=2∠1
∠A+∠C=∠3+∠2+∠4
=2∠1+∠4
Also ∠1=∠5
∠A+∠C=∠1+∠4+∠5=180°
Similarly, we can show that ㄥB+ㄥD=180°
Hence, the opposite angles of an isosceles trapezium are supplementary.

Given: ABCD is a parallelogram and line segments AX, CY bisect the angles A and C respectively.
To Prove: AX IICY.
Proof: \(\because\) ABCD is a parallelogram.
\(\therefore\) \(\angle \)A = \(\angle \)C  I Opposite \(\angle \)s of a parallelogram are equal

⇒  \(1\over 2\)\(\angle \)A=\(1\over 2\)\(\angle \)C     |\(\because \)Halves of equals are equal
⇒    \(\angle \)1 = \(\angle \)2    ...(1)      I    \(\because\)   AX is the bisector of \(\angle \)A and CY is the bisector of \(\angle \)C Now, AB II DC and CY intersects them
 \(\therefore\)  \(\angle \)2 =\(\angle \)3  ....(2)   I Alternate interior \(\angle \) s 
From (1) and (2), we get
\(\angle \)1 = \(\angle \)3
But these form a pair of equal corresponding angles
\(\therefore\)  AX II CY.

(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) Since sum of adjacent angles of a parallelogram is 180°
ஃ We have ㄥA+ㄥB=180
⇒ 5x + 7 + 3x - 3 = 180
⇒ 8x + 4 = 180
⇒ 8x = 176
⇒ x=\(\frac{176}{8}\)=22

ㄥA=(5x+7)°=(5x22+7)=117°
ㄥB=(3x-3)°=(3x22-3)=63°
ㄥC=ㄥA=117°
and ㄥD=ㄥA=63°
(ii) Properties of parallelogram.
(iii) Energy conservation is necessary for a happy and prosperous future.

ㄥR=80° (Given)
SR II PQ and RQ is a transversal
ஃ ㄥR+ㄥQ=180°
ㄥQ=180°-80°
= 100°
Similarly, ㄥQ+ㄥP=180°
⇒ ㄥP=180°-100°=80°
and ㄥS+ㄥR=180°
⇒ ㄥS=180°-80°=100°
Hence, ㄥP=80°, ㄥQ=100°, ㄥR=80°, ㄥS=100°

(ii) Property of co-interior angles when a pair of straight lines intersected by another straight line (Geometry)
(iii) Diligence i.e., dedication, determination, and hard work.

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