# 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(ΔABC) = ar(ΔDBC)

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.
ar(||gm EACB)= ar(||gm DFCB)...(i)
Now AB is the diagonal of ||gm EACB
ar(
ΔEAB) = ar(ΔABC)
ar(
ΔABC) = (1/2)ar(||gm EACB) .....(ii)
Similarly,
ar(
ΔDBC) = (1/2) ar(||gm DFCB)...(iii)
From equations (i), (ii) and (iii), we get
ar(Δ ABC) = ar(Δ DBC)

Corollary 1
The area of a triangle is half the product of any of its side and the corresponding altitude.
Given: In
ΔABC, AE is the altitude to the base BC
To Prove: ar(
ΔABC) = × BC × AE Construction: Through A and C, draw lines parallel respectively to BC and AB intersecting each other at D.
Proof: Since AB
ïï DC and AD ïï BC, ABCD is a parallelogram and AC is the diagonal.
ar(
ΔABC)   = ar(ïïgm ABCD) ...(i)
BC is the side of
ïïgm ABCD and AE is the corresponding altitude
...       ar(
ïïgm ABCD) = BC × AE ...(ii)
From equations (i) and (ii), we get
ar(
ΔABC) = BC × AE

Example :

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

Solution :
 Median of a triangle divides it into two triangles of equal area Let AD be the median of ΔABC. We have to show that ar(ΔABD) = ar(ΔADC).
Draw AE
BC, meeting BC in E ar(ΔABD) = BD x AE ...(i)
Again ar(
ΔADC) = DC x AE ...(ii)
Also, BD = DC ...(iii)...(D is the mid-point of BC)
From (i), (ii) and (iii), we get ar(
II Method:
ΔABD and ΔADC are on equal bases and between same parallels BC and AX.
... ar(    