Triangles on the Same Base and between the Same Parallels
Theorem 2
Triangles on the same base and between same parallels are equal in area.
Given : Two triangles ABC and DBC are on the same base BC and between same parallels EF and BC.
To Prove : ar(Î”ABC) = ar(Î”DBC)
Construction : Through B, draw BE  AC intersecting line AD in E and through C draw CF  BD intersecting the line DA in F.
Proof : EACB and DFCB are parallelograms (Since two pairs of opposite sides are parallel)
Also gm EACB and gm DFCB are on the same base BC and between same parallels EF and BC.
âˆ´ ar(gm EACB)= ar(gm DFCB)...(i)
Now AB is the diagonal of gm EACB
âˆ´ ar(Î”EAB) = ar(Î”ABC)
âˆ´ ar(Î”ABC) = (1/2)ar(gm EACB) .....(ii)
Similarly,
ar(Î”DBC) = (1/2) ar(gm DFCB)...(iii)
From equations (i), (ii) and (iii), we get
ar(Î” ABC) = ar(Î” DBC)
Corollary 1
The area of a triangle is half the product of any of its side and the corresponding altitude.
Given: In Î”ABC, AE is the altitude to the base BC
To Prove: ar(Î”ABC) = Ã— BC Ã— AE
Construction: Through A and C, draw lines parallel respectively to BC and AB intersecting each other at D.
Proof: Since AB Ã¯Ã¯ DC and AD Ã¯Ã¯ BC, ABCD is a parallelogram and AC is the diagonal.
ar(Î”ABC) = ar(Ã¯Ã¯^{gm } ABCD) ...(i)
BC is the side of Ã¯Ã¯^{gm }ABCD and AE is the corresponding altitude
.^{.}. ar(Ã¯Ã¯^{gm }ABCD) = BC Ã— AE ...(ii)
From equations (i) and (ii), we get
ar(Î”ABC) = BC Ã— AE
Show that a median of a triangle divides it into two triangles of equal area.
Median of a triangle divides it into two triangles of equal area 

Let AD be the median of Î”ABC. We have to show that ar(Î”ABD) = ar(Î”ADC).
Draw AE âŠ¥ BC, meeting BC in E ar(Î”ABD) = BD x AE ...(i)
Again ar(Î”ADC) = DC x AE ...(ii)
Also, BD = DC ...(iii)...(D is the midpoint of BC)
From (i), (ii) and (iii), we get ar(Î”ABD) = ar(Î”ADC)
II Method: Î”ABD and Î”ADC are on equal bases and between same parallels BC and AX.
.^{.}. ar(Î”ABD) = ar(Î”ADC)