Let ABC and A'BC be two triangles on the same base BC and between the same parallels PQ and RS.
 
Draw \(BM\bot PQ\),then,
\(ar(\Delta ABC)\)
 \( =\frac { 1 }{ 2 } \times Base\times Corresponding\ altitude\)
\(\\ =\frac { 1 }{ 2 } \times BC\times BM\quad \quad \quad \quad .....(i)\)
\(\\ ar(\Delta A'BC)\)
\(\\ =\frac { 1 }{ 2 } \times Base\times Corresponding\ altitude\)
\(\\ =\frac { 1 }{ 2 } \times BC\times BM\quad\quad \quad \quad \quad .....(ii)\)
Form (i) and (ii),we get
 \(ar(\Delta ABC)=ar(\Delta ABC)\)
Hence, \(\Delta ABC\) and \(\Delta A'BC\)  are equal in area.

Given: Diagonals AC and BD of a trapezium ABCD with AB || CD intersect each other at O.
To Prove: ar(\(\Delta \)AOD) = ar(\(\Delta \) BOC)
Proof:  \(\Delta \)ADB and \(\Delta \)ACB are on the same base AB and between the same parallels AB and DC
ar(\(\Delta \)ADB) = ar(\(\Delta \) ACB)
 |Two triangles on the same base and between the same parallels are equal in area
\(\Rightarrow \)  ar(\(\Delta \) ADB) - ar(\(\Delta \) AOB)
\(\Rightarrow \)  ar(\(\Delta \) ACB) - ar(\(\Delta \)AOB)
I Subtracting ar( \(\Delta \)AOB) from both sides
\(\Rightarrow \)  ar(\(\Delta \) AOD) = ar(\(\Delta \)BOC)

Given: AB || DC in the given figure.
To Prove: ar(BDE) = ar(ACED)
Proof:\(\Delta \)  ADC and \(\Delta \) BDC are on the same base DC and between the same parallels AB and DC
ar(\(\Delta \) ADC) = ar(\(\Delta \) BDC)
\(\Rightarrow ar(\Delta \ ADC)+ar(\Delta \ DCE)\)
\( =ar(\Delta \ BDC)+ar(\Delta \ DCE)\)
|Adding ar(\(\Delta \quad DCE\)) to both sides
\(\Rightarrow \ ar(ACED)=ar(BDE)\)
\(\\ \Rightarrow \ ar(BDE)\ =ar(ACED)\)

Given: ABCD is a quadrilateral and BD is one of its diagonals.
To Prove: ABCD is a parallelogram and to determine its area.
Proof:\(\angle ABD=\angle BDC(=90°)\)        |Given
But these angles form a pair of equal alternate.interior angles for lines AB, DC and a transversal BD
AB || DC
Also,  AD = DC (= 3 cm) I Given
Hence, quadrilateral ABCD is a parallelogram.
I A quadrilateral is a parallelogram if its one pair of opposite sides are parallel and equal
Now,
 \(ar(||gm\ ABCD)=base\times Corresponding\ altitude\)
\(\\ =3\times 4\)
\(\\ =12{ cm }^{ 2 }\)

Given: BE = 14 cm, AD = 8 cm
\(\therefore \ Area(\triangle ADB)=\frac { 1 }{ 2 } \times 8\times 14\)
= 56 cm²
\(\because\) ABCD is a parallelogram.
\(\therefore\) ar(\(\Delta \)DBC) = ar(\(\Delta \)ADB) = 56 cm².

A diagonal of a parallelogram divides it into two triangles of equal area
\(\therefore\) ar(\(\Delta\)ABD)=  ar(\(\Delta\)BCD)
Since, given, ar(\(\Delta\)ABD) = 8 cm²
\(\therefore\) ar(\(\Delta\)BCD) = 8 cm².

12 cm²

7.5 cm²

6.4cm

AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas.
∴ Area (ΔABD) = Area (ΔACD)
⇒ Area (ΔABD) = 1/2Area (ΔABC)... (1)
In ΔABD, E is the mid-point of AD. Therefore, BE is the median.
∴ Area (ΔBED) = Area (ΔABE)
⇒ Area (ΔBED) = 1/2Area (ΔABD)
⇒ Area (ΔBED) = 1/2 x 1/2Area (ΔABC) [From equation (1)]
⇒ Area (ΔBED) = 1/4Area (ΔABC)

A median of a triangle divides it into two triangles of equal area.

A median of a triangle divides it into two triangles of equal areas

A diagonal of a parallelogram divides it into two congruent triangles. Two congruent triangles have equal areas. The diagonals of a parallelogram bisect each other.A median of a triangle divides it into two triangles of equal areas.

Theorem

\(\Delta\)TQR and parallelogram SRQP lie on the same base RQ and between the same parallels RQ and SP.
\(\Delta\)PDC and trapezium ABCD lie on the same base DC and between the same parallels AB and De.

Diagonal=\(\sqrt { 2 }\) side;Area=(side)²

Theorem