Ninth Grade Mathematics: Triangles on Same Base & Between Parallels
Just as parallelograms on the same base and between the same parallels possess identical areas, triangles sharing a base between parallel boundaries obey the same elegant geometric invariance. In this chapter, you will master the fundamental theorems connecting triangle areas to parallelograms, prove the equality of triangular areas, and explore the converse theorem used to establish parallelism.
The Relationship Between Triangle and Parallelogram Areas
A triangle is fundamentally half of a parallelogram. Recall that a diagonal divides any parallelogram into two congruent triangles, and congruent figures have equal areas.
The Parallelogram-Triangle Area Theorem:
If a triangle and a parallelogram lie on the same base and between
the same parallels, then the area of the triangle is equal to half
the area of the parallelogram.
ar(Triangle ABC) = (1/2) * ar(Parallelogram ABED)
D C E (Line DE || Base AB)
*-------------------*---------*
/ \ / /
/ \ / /
/ \ / /
/ \ / /
*---------*---------*---------*
A B
ar(Triangle ABC) = (1/2) * ar(Parallelogram ABED)
Since the area of a parallelogram is Base * Altitude, this theorem immediately yields the universal formula for the area of any triangle:
Area of a Triangle = (1/2) * Base * Altitude = (1/2) * b * h
Triangles on the Same Base and Between the Same Parallels
From the relation to parallelograms, the primary theorem of triangular areas follows directly:
The Triangle Area Theorem:
Triangles on the same base (or equal bases) and between the same parallels
are equal in area.
In the diagram below, if Line l is parallel to Base BC:
ar(Triangle ABC) = ar(Triangle DBC)
Line l (Parallel to Base BC)
A D
---------*-----------------*---------
/ \ / \
/ \ / \
/ \ / \
/ \ / \
*---------*-------*---------*----
B C B C
ar(Triangle ABC) == ar(Triangle DBC)
Proof: Through vertex A, draw line AP parallel to BC meeting the ceiling parallel. Through vertex D, draw line DQ parallel to BC. This forms two parallelograms on the same base BC between the same parallels. By the Parallelogram Area Theorem, ar(Parallelogram 1) = ar(Parallelogram 2). Since each triangle has area equal to half of its respective parallelogram: ar(Triangle ABC) = (1/2)ar(Parallelogram 1) = (1/2)ar(Parallelogram 2) = ar(Triangle DBC).
The Converse Theorem: Proving Lines Parallel
The converse theorem provides an indispensable tool for proving that two lines are parallel:
Converse Theorem:
Two triangles having the same base (or equal bases) and equal areas,
lying on the same side of the base, lie between the same parallels.
If ar(Triangle ABC) == ar(Triangle DBC) on base BC:
THEN Line AD is PARALLEL to Line BC (AD || BC).
Proof of Converse by Contradiction:
1. Let Triangle ABC and Triangle DBC have equal areas on base BC.
2. Assume, for contradiction, that AD is not parallel to BC.
3. Draw a line through A parallel to BC, intersecting line segment BD (or CD)
at a different point E.
4. Join EC to form Triangle EBC.
5. Since Triangle ABC and Triangle EBC lie on the same base BC and between
parallels, ar(Triangle ABC) = ar(Triangle EBC).
6. But we were given ar(Triangle ABC) = ar(Triangle DBC).
7. Therefore, ar(Triangle EBC) = ar(Triangle DBC).
8. This implies that a part equals the whole, which is impossible unless
point E coincides with point D!
9. Therefore, line AD must be parallel to line BC.
Trapezoids and Non-Parallel Side Triangles
An extraordinary property of every trapezoid involves the triangles formed by intersecting diagonals:
Theorem:
In any trapezoid ABCD with AB || CD, if diagonals AC and BD intersect at O,
then the areas of the two triangles formed on the non-parallel sides are EQUAL:
ar(Triangle AOD) = ar(Triangle BOC)
D C
*--------------------*
/ \ / \
/ \ O / \
/ \ * / \
/ \ / \ / \
/ AOD \ / \ / BOC \
*-----------*-----*--*-----------*
A \ / B
\ / AOB
Proof: Triangles ABC and ABD share the common base AB and lie between parallels AB and CD, so ar(Triangle ABC) = ar(Triangle ABD). Both triangles contain triangle AOB. By the Area Addition Axiom, subtract ar(Triangle AOB) from both: ar(Triangle ABD) - ar(Triangle AOB) = ar(Triangle ABC) - ar(Triangle AOB). This leaves ar(Triangle AOD) = ar(Triangle BOC).
Chapter Exercises
Exercise 1: In a triangle ABC, point D lies on side BC such that BD = DC (AD is a median). Prove that the area of triangle ABD is equal to the area of triangle ACD.
Exercise 2: The diagonals of quadrilateral ABCD intersect at point O. If ar(Triangle AOD) = ar(Triangle BOC), prove that ABCD is a trapezoid.
Exercise 3: Triangle ABC and parallelogram ABDE lie on the common base AB and between the same parallel lines AB and CE. If the area of triangle ABC is 28 square cm, find the area of parallelogram ABDE.
Solutions and Explanations
Solution 1: Let AD be the median of triangle ABC, so D is the midpoint of base BC, meaning BD = CD. Draw the perpendicular altitude AM from vertex A to the line containing base BC. The altitude AM is common to both triangle ABD and triangle ACD because both triangles share vertex A and their bases lie on the same straight line BC. The area of triangle ABD is (1/2) * base * height = (1/2) * BD * AM. The area of triangle ACD is (1/2) * CD * AM. Since BD = CD, substituting yields (1/2) * BD * AM = (1/2) * CD * AM. Therefore, ar(Triangle ABD) = ar(Triangle ACD). A median always bisects a triangle into two triangles of equal area.
Solution 2: We are given that ar(Triangle AOD) = ar(Triangle BOC). Add the area of triangle AOB to both sides of this equation: ar(Triangle AOD) + ar(Triangle AOB) = ar(Triangle BOC) + ar(Triangle AOB). By the Area Addition Axiom, ar(Triangle AOD) + ar(Triangle AOB) = ar(Triangle ABD), and ar(Triangle BOC) + ar(Triangle AOB) = ar(Triangle ABC). Therefore, ar(Triangle ABD) = ar(Triangle ABC). Notice that Triangle ABD and Triangle ABC lie on the common base AB and have equal areas, and lie on the same side of AB. By the Converse of the Triangle Area Theorem, these two triangles must lie between the same parallels, which proves that line AB is parallel to line CD (AB || CD). Since quadrilateral ABCD has a pair of parallel opposite sides, ABCD is a trapezoid.
Solution 3: By the Parallelogram-Triangle Area Theorem, when a triangle and a parallelogram share the same base and lie between the same parallel lines, the area of the triangle is exactly half the area of the parallelogram: ar(Triangle ABC) = (1/2) * ar(Parallelogram ABDE). Multiplying both sides by 2 yields ar(Parallelogram ABDE) = 2 * ar(Triangle ABC). Given ar(Triangle ABC) = 28 square cm: ar(Parallelogram ABDE) = 2 * 28 = 56 square cm.