Here ㄥ1=ㄥ2 (AE is the angle bisector)
But ㄥ1=ㄥ3 (alternate angles as AD||BC)
ㄥ3=ㄥ2

Hence BE=AB (sides opposite to equal angles)
But BE=\(\frac{1}{2}\) BC, (E is the mid-point of BC)
AB=\(\frac{1}{2}\)BC
and BC=AD (opposite sides of a parallelogram)
AB=\(\frac{1}{2}\)AD
Hence Proved.

A parallelogram PQRS in which the bisectors of LP and LQ meet SR at O.
To Prove: ㄥPOQ=90°
Now, since PQRS is a parallelogram. Therefore,
PSIIQR
Now, PS II QR and transversal PQ intersects them.
ㄥP+ㄥQ=180°
(∵ Sum of consecutive interior angles is 180°)
\(\frac{1}{2}\)ㄥP+\(\frac{1}{2}\)ㄥQ=90°
⇒ ㄥ1+ㄥ2=90° ( OP is bisector of ㄥP...(i) and OQ is bisector of ㄥQ.
ㄥ1=\(\frac{1}{2}\) ㄥP and ㄥ2=\(\frac{1}{2}\) ㄥQ)
Now, in ΔPOQ
ㄥ1+ㄥPOQ+ㄥ2=180°
⇒ 90°+ㄥPOQ=180°
⇒ ㄥPOQ=90°

In Δs APD and CQB
AD = BC (Opp. sides of a parallelogram)
PD = BQ (Given)
∠ADP=∠QBC
⇒ ΔAPD≡ΔCQB
⇒ AP = CQ (c.p.c.t)
In Δs AQB and CPD
AB = DC, BQ = DP
and ∠AQB=∠PDC
ஃ ΔAQB≡ΔCPB ⇒ AQ=CP
(iii) In quad. APCQ, 
AP = CQ and AQ = CP
⇒ APCQ is a parallelogram.

In ΔPSQ and ΔRQS
PQ = SR (= 4cm)
SQ = SQ (common)
ㄥSQP =∠QSR (Each 90°)
ΔPSQ ≡ ΔRQS by SAS rule.
Thus, ar (ΔPSQ) = ar(ΔSRQ)
[∵ Areas of congruent figures are equal.]

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.

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.

Two triangles ABC and DEF, such that

AB = DE and AB II DE
Also BC = EF and BC II EF
Proof: (i) In a quadrilateral ABED,
AB = DE and AB II DE
⇒ One pair of opposite sides are equal and parallel.
ABED is a parallelogram
⇒ AD = BE and AD II BE....(i)
(ii) In quadrilateral BCFE,
BC = EF and BC II EF
⇒ One pair of opposite sides are equal and parallel.
BCFE is a parallelogram
CF = BE and CF II BE ...(ii)
(iii) From equations (i) and (ii), we get
AD = CF and AD II CF
⇒ ACFD is a parallelogram
AC = DF andAC II DF
(iv) In ∆ABC and ∆DEF
AB = DE (Given)
BC = EF (Given)
and AC = DF (Proved above in part (c))
So by S.S.s.,ΔABC≅ΔDEF.

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.

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°

∠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

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°

10.5 cm

5 cm

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

Theorm

no of angle of a rhombus is 90⁰

opposite sides are not equal in a kite