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: 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.

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°

ㄥADC+ㄥBCD=180°
⇒ \(\frac{1}{2}\)ㄥADC+\(\frac{1}{2}\)ㄥBCD=90°
or ㄥ1+ㄥ2=90°

In ΔODC,
ㄥ1+ㄥ2+ㄥDOC=180°
ㄥDOC=90°

∠APR=ㄥDRP
or ㄥ1=ㄥ2
But these are alternate interior angles
SP II RQ, SR II PQ
PQRS is a parallelogram
∠APR+ㄥBPR=180°,(linear pair)
⇒ \(\frac{1}{2}\) ∠APR+\(\frac{1}{2}\)ㄥBPR=\(\frac{1}{2}\)x180°
⇒ ∠1+ㄥ3=90°
⇒ ∠SPQ=90°

PQRS is a rectangle

72°, 108°, 72°, 108°

\(36°\),\(144°\),\(36°\), \(144°\)

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.

PQ II DB, SR II DB
 
\(\therefore\) PQ II SR
Similarly, PS IIQR
\(\therefore\) PQRS is a parallelogram.

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.

Theorem

\(\angle \)AGF +\(\angle \) GAE = \(​​180°\)
\(\Rightarrow \) \(\angle \)AGF+\(​​60°\)=\(​​180°\)
\(\Rightarrow \)\(\angle \)AGF=\(​​120°\)

Theorem

see a square

See a blackboard

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

opposite sides are not equal in a kite

(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.