55°

Given: ABCD is a trapezium in which AB II DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F.
To Prove: F is the mid-point of Be.
Proof: Let DB intersect EF at G.
In \(\Delta \)DAB,
\(\because\) E is the mid-point of DA and EG II AB
\(\therefore\) G is the mid-point of DB     | By converse of mid-point theorem
Again, in  \(\Delta \)BDC, 
 \(\because\)  G is the mid-point of BD and GF II AB II DC
 \(\therefore\)  F is the mid-point of BC.   | By converse of mid-point theorem

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.

Given: In parallelogram ABCD, AC = BD   To Prove: ||gm ABCD is a rectangle.

Proof: In \(\Delta\) ACB and \(\Delta\)BDA,
AC = BD                                 I Given
AB = BA                                 I Common
BC = AD                                 I Opposite sides of II gm ABCD
\(\therefore\) \(\Delta\)ACB =\(\Delta\)BDA               | SSS Congruence Rule
\(\therefore\)   \(\angle ABC=\angle BAD\quad \quad \quad ......(1)\quad C.P.C.T\) 
AD || BC                    | Opp. sides of II gm ABCD
and transversal AB intersects them.
 \(\angle BAD+\angle ABC=180°\quad \quad \quad ....(2)\)
|Sum of consecutive interior angles on the same side of a transversal is 180
From (1) and (2),
 \(\angle BAD=\angle ABC=90°\)
\(\therefore\) ​​​​​|| gm ABCD is a rectangle.

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°

Since opposite angles of a parallelogram are equal
3x - 2 = 63 - 2x
⇒ x=13°
Angles of a parallelogram :
(39 - 2)°, (180- 37)°, (63 - 26)°, (180- 37)°
i.e., 37°,143°,37°,143°.

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

72°, 108°, 72°, 108°

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.

ΔPTS and ΔPXS lie on same base and between the same parallels XT and PS
ar (PTS) = ar (PXS) ...(i)
Also, ΔSQR and ΔSQY lie on same base SQ and between the same parallels SQ and RY
ar (SQR) = ar (SQY) ...(ii)
Adding (i) & (ii), we get,
ar (PTS) + ar (SQR) = ar (PXS) + ar (SQY)
...(iii)
Adding ar (PQS) on both sides of (3), we get
ar (PTS) + ar (PQS) + ar (SQR)
= ar (PXS) + ar (PQS) + ar (SQY)
i.e., ar (PQRST) = ar (SXY)
Hence proved.

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: 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

(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