In \(\Delta\)GHK; F and E are the mid-points of HG and GK respectively.
\(\therefore\) By mid-point theorem,
\(FE=\frac { 1 }{ 2 } KH\quad and\quad FE||KH\)  ...........(i)
In quadrilateral EFHK,
EF || HK        (by (i))
\(\therefore\) EFHK is a trapezium.
Also,
ar(EFHK) = ar(FHD) + ar(DEF) + ar(DEK)   .............(ii)
We have,    FE || HD and FE = HD
\(\therefore\) FEDH is a ||^gm,
So, ar(FHD) = ar(DEF)   ..........(iii)
Similarly, DFGE is a ||^gm, 
\(\therefore\) ar(DEF) = ar(GEF)   ...........(iv)
Also, DFEK is a ||^gm,
\(\therefore\) ar(DEF) = ar(DEK)    ...........(v)
Using (iii), (iv) & (v) we get,
ar(GEF) = ar(FHD)
= ar(DEK) = ar(DEF)
\(=\frac { 1 }{ 4 } \) ar(GHK)                ...........(vi)
Using (vi) and (ii), we get
ar(EFHK) = \(\frac { 3 }{ 4 } ar(GHK)\)
Hence proved

Given: BQ||CR
\(\because\) \(\Delta \) BCQ and \(\Delta \)BQR are on the same base BQ and between the same parallel BQ and CR.
\(\therefore\) ar(\(\Delta \)BCQ) = ar (\(\Delta \)BQR)    ......(i)
Also,  AP||BQ.     (Given)

Again, \(\Delta \)ABQ and \(\Delta \)PBQ are on the same base BQ and between the same parallels BQ and AP.
\(\therefore\) ar(\(\Delta \)ABQ) = ar (\(\Delta \)PBQ)    .......(ii)
Adding (1) and (2), we get
ar(\(\Delta \)BCQ) + ar(\(\Delta \)ABQ) = ar(\(\Delta \)BQR) + ar(\(\Delta \)PBQ)
ar(\(\Delta \)AQC) = ar(\(\Delta \)PBR).

ar(\(\Delta\)DRC) = ar(\(\Delta\)DPC),   (Given)
But they are on the same base DC.
Therefore, \(\Delta\)DRC and \(\Delta\)DPC must lie between the same parallels.
So, DC || RP
i.e., one pair of opposite sides of quadrilateral DCPR is parallel.
therefore, DCPR is a trapezium.
Also, ar(\(\Delta\)BDP) = ar(\(\Delta\)ARC)    (Given)      ........(i)
and ar (\(\Delta\)DPC) = ar(\(\Delta\)DRC) (Given) .........(ii)
Subtracting (ii) from (i), we get
ar(\(\Delta\) BDP) - ar(\(\Delta\)DPC) = ar(\(\Delta\)ARC) - ar(\(\Delta\)DRC) ar(\(\Delta\)BDC) = ar(\(\Delta\)ADC)

But they are on the same base DC.
Therefore, \(\Delta\)BDC and \(\Delta\)ADC must lie between the same parallels.
So, AB || DC
i.e., One pair of opposite sides of quadrilateral ABCD is parallels.
So, AB || DC
i.e., One pair of opposite sides of quadrilateral ABCD is parallel.
Therefor, ABCD is a trapezium.

Given: \(\Delta\)ABQ and parallelogram ABCD are on the same base AB and between the same parallels DC and AB.
To prove:
Area(\(\Delta\)ABQ) = \(\frac { 1 }{ 2 } \)Area (Parallelogram ABCD)
Construction: Extend DC to R so that BR||AQ
Proof: DCBA and QRBA are on the same base and between same parallels
ar(DCBA) = ar(QRBA)    ............(i)
A diagonal divides a parallelogram into two congruent triangles with equal area
\(\Rightarrow \ ar(QAB)=\frac { 1 }{ 2 } ar(QCBA)\)     .....(ii)
From (i) and (ii)
ar(QAB) = \(\frac { 1 }{ 2 } ar\left( DCBA \right) \)

Given: A parallelogram MNOP is which one of the diagonals is PN

To show: ar(\(\Delta\)PMN)=ar(\(\Delta\)PON)
Now, since two congurent figures are equal in area, so we will show that
\(\Delta\)PMN\(\cong \)\(\Delta\)PON
\(\therefore\) In \(\Delta\)'s PMN and PON,
MN = PO
[MNOP is a ||^gm, \(\therefore\) opp.sides are equal MN=PO
PM = ON
and, PN = NP
\(\therefore\) By SSS congurence criterion, we get
\(\Delta\)PMN\(\cong \)\(\Delta\)PON
Hence ar(\(\Delta\)PMN) = ar(\(\Delta\)PON)

Given: Diagonals AC and BD of a quadrilateral ABCD intersect at 0 in such a way that ar(\(\Delta \) AOD) = ar(\(\Delta \) BOC).

To Prove: \(\Box \) ABCD is a trapezium.
Proof: ar(\(\Delta \) AOD) = ar(\(\Delta \) BOC) IGiven
ar(\(\Delta \) AOD) + ar(\(\Delta \) AOB)
= ar(\(\Delta \) BOC) + ar(\(\Delta \) AOB)
I Adding ar(\(\Delta \)AOB) to both sides
= ar(\(\Delta \) ABD) = ar(\(\Delta \)ABC)
But \(\Delta \)ABD amd \(\Delta \) ABC are on the same base AB.
\(\Delta \) ABD and \(\Delta \) ABC will have equal corresponding altitudes.
\(\Delta \) ABD and \(\Delta \) ABC will lie between the same parallels.
AB || DC
ABCD is a trapezium.
A quadrilateral is a trapezium if exactly one pair of opposite sides are parallel

Given: AP || BQ || CR.
To Prove: ar(\(\Delta \) AQC) = ar(\(\Delta \)PBR).
Proof: \(\Delta \) BAQ and \(\Delta \)BPQ are on the same base BQ and between the same parallels BQ and AP.
ar(\(\Delta \) BAQ) = ar(\(\Delta \) BPQ) ...(1)
|Two triangles on the same base (or equal bases) and between the same parallels are equal in area
 \(\Delta \) BCQ and \(\Delta \)BQR are on the same base BQ and between the same parallels BQ and CR.
ar(\(\Delta \) BCQ) = ar(\(\Delta \) BQR) ...(1)
|Two triangles on the same base (or equal bases) and between the same parallels are equal in area
Adding the corresponding sides of (1) and (2), we get
ar(\(\Delta \) BAQ) + ar(\(\Delta \) BCQ)
= ar(\(\Delta \) BPQ) + ar(\(\Delta \) BQR)
= ar(\(\Delta \) AQC) = ar(\(\Delta \) PBR).​​​​

Given: AB CD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y.

To Prove: ar(\(\Delta \)ADX) = ar(\(\Delta \)ACY).
Construction: Join CX.
Proof: \(\Delta \)ADX and \(\Delta \)ACX are on the same base AX and between the same parallels AB and DC.
ar(\(\Delta \)ADX) = ar(\(\Delta \)ACX)        ...(1)
|Two triangles on the same base (or equal bases) and between the same parallels are equal in area
\(\Delta \)ACX and \(\Delta \)ACY are on the same base AC and between the same parallels AC and XY.
ar(\(\Delta \)ACX) = ar(\(\Delta \)ACY)        ...(2)
|Two triangles on the same base (or equal bases) and between the same parallels are equal in area
From (1) and (2), we get
ar(\(\Delta \)ADX) = ar(\(\Delta \)ACY).

Let ABCD be the plot of land of the shape of a quadrilateral. Let the portion ADE be taken over by the Gram Panchayat of the village from one corner D to construct a Health Centre. Join AC. Draw a line through D parallel to AC to meet BC produced in P. Join EP. Then Itwaari must be given the land ECP adjoining his plot so as to form a triangular plot ABP as then
\(ar(\Delta ADE)=ar(\Delta PEC)\)
  
Proof: \(\Delta \)DAP and \(\Delta \)DCP are on the same base DP and between the same parallels DP and AC.
ar(\(\Delta \)DAP) = ar(\(\Delta \)DCP)
| Two triangles on the same base (or equal bases) and between the same parallels are equal in area
ar(\(\Delta \)DAP) - ar(\(\Delta \)DEP)
= ar(\(\Delta \)DCP) - ar(\(\Delta \)DEP)
I Subtracting ar(\(\Delta \) DEP) from both sides
ar(\(\Delta \) ADE) = ar(\(\Delta \) PCE)
ar(\(\Delta \)DAE) + ar(\(\Box \) ABCE)
= ar(\(\Delta \)PCE) + ar(\(\Box \) ABCE)
I Adding are \(\Box \) ABCE) to both sides
ar(\(\Box \) ABCD) = ar(\(\Delta \) ABP).

ar(parallelogram ABCD) = AB x AE
= 16 x 8 cm²
= 128 cm²             ...(1)
ar(parallelogram ABCD) = AD x CF
= AD x 10 cm²   .......(2)
From (1) and (2), we get
AD x 10 = 128
AD = \(\frac { 128 }{ 10 } \)
AD = 12.8 cm.