### Theory:

Theorem I: Area of a triangle is equal to half the product of its base and the corresponding altitude.

Given: $$PQRS$$ is a parallelogram, and $$PR$$ is one of its diagonal.

Construction: Draw $$PX \perp SR$$.

To prove: $$ar(\triangle PSR) = \frac{1}{2} \times SR \times PX$$

Proof: Since $$\triangle PRS$$ and parallelogram $$PQRS$$ are on the same $$SR$$ and between the same parallels, then we have:

$$ar(PRS) = \frac{1}{2} \times ar(PQRS)$$

$$ar(PRS) = \frac{1}{2} \times SR \times PX$$ [Area of a parallelogram $$=$$ base $$\times$$ height]

Therefore, the area of a triangle is equal to half the product of its base and the corresponding altitude.

Hence proved.
Theorem II: Two triangles on the same base and between the same parallels are equal in area.

Given: $$\triangle ABC$$ and $$\triangle DBC$$ are on the same base $$BC$$ and between the same parallels $$BC$$ and $$XY$$.

To prove: $$ar(\triangle ABC) = ar(\triangle DBC)$$

Construction: Draw $$BP \parallel AC$$ and $$CQ \parallel BD$$ meeting $$XY$$ at $$P$$ and $$Q$$.

Proof: From the figure, we can see that $$BCAP$$ and $$BCQD$$ are parallelogram and are on the same base $$BC$$ and between the same parallels $$BC$$ and $$XY$$.

$$ar(BCAP) = ar(BCQD)$$ ---- ($$1$$) [Parallelograms on the same base and between the same parallels are equal in the area]

Also, $$\triangle ABC$$ and parallelogram $$BCAP$$ are on the same base $$BC$$ and between the same parallels $$BC$$ and $$XY$$.

$$ar(\triangle ABC) = \frac{1}{2} \times ar(BCAP)$$ ---- ($$2$$) [If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of a parallelogram]

Similarly, $$\triangle DBC$$ and parallelogram $$BCQD$$ are on the same base $$BC$$ and between the same parallels $$BC$$ and $$XY$$.

$$ar(\triangle DBC) = \frac{1}{2} \times ar(BCQD)$$ ---- ($$3$$) [If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of a parallelogram]

Substituting the value of equations ($$2$$) and ($$3$$) in ($$1$$), we have:

$$2 \times ar(\triangle ABC) = 2 \times ar(\triangle DBC)$$

$$ar(\triangle ABC) = ar(\triangle DBC)$$

Hence proved.
Theorem III: Two triangles having the same base and equal areas lie between the same parallels.

Given: $$PRS$$ and $$QRS$$ are two triangles on the same base $$SR$$ with equal areas.

That is, $$ar(\triangle PRS) = ar(\triangle QRS)$$

To prove: The two triangles lie between the same parallels. That is, $$PQ \parallel SR$$

Construction: Join $$PQ$$. Draw $$PX \perp SR$$ and $$QY \perp SR$$

Proof: Since $$PX \perp SR$$ and $$QY \perp SR$$, then

$$PX \parallel QY$$ ---- ($$1$$) [Lines perpendicular to the same line are parallel to each other]

Also, we are given that $$ar(\triangle PRS) = ar(\triangle QRS)$$

$$\frac{1}{2} \times RS \times PX = \frac{1}{2} \times RS \times QY$$

$$PX = QY$$ ---- ($$2$$)

Now, consider $$PQXY$$,

$$PX \parallel QY$$ and $$PX = QY$$ [From equations ($$1$$) and ($$2$$)]

Therefore, $$PQXY$$ is a parallelogram. Because a pair of opposite sides are equal and parallel.

Thus, $$PQ \parallel XY$$ [Opposite sides of a parallelogram are parallel]

$$PQ \parallel RS$$ [Since $$XY$$ lies on the line $$RS$$]

Therefore, two triangles having the same base and equal areas lie between the same parallels.

Hence proved.