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Chapter 9 Areas Of Parallelograms And Triangles (Concepts)
Welcome to this insightful chapter dedicated to exploring the concept of Area, but with a unique focus that goes beyond simple formula application. Here, we delve into the fundamental geometric relationships governing the areas of parallelograms and triangles, particularly when they share common bases and are confined between the same parallel lines. This chapter emphasizes geometric reasoning and comparison of areas based on spatial configuration, often allowing us to deduce equality or specific ratios without needing explicit numerical calculations of base and height.
We begin by reinforcing the basic understanding of area as the measure of the region enclosed by a plane figure. A crucial preliminary concept is the relationship between congruence and area: it's established that congruent figures (figures with identical shape and size) necessarily have equal areas. However, the converse statement is not always true – two figures can possess the exact same area without being congruent (e.g., a $4 \times 4$ square and an $8 \times 2$ rectangle both have an area of 16 square units but are clearly not congruent).
The first cornerstone theorem of this chapter establishes a profound relationship between parallelograms sharing specific geometric constraints: Parallelograms on the same base (or on equal bases) and lying between the same parallel lines are equal in area. The proof of this theorem typically involves cleverly dissecting the parallelograms and demonstrating the congruence of certain triangular parts, ultimately showing that the areas must be identical. This theorem is powerful because it allows us to equate the areas of potentially different-looking parallelograms based solely on their shared base and parallel confinement, independent of their specific angles or side lengths (other than the base).
Building directly upon this, the chapter investigates the connection between the area of a triangle and that of a related parallelogram. The key theorem states: If a triangle and a parallelogram stand on the same base and lie between the same parallel lines, then the area of the triangle is exactly half the area of the parallelogram. This is often visualized by considering the parallelogram formed by completing the triangle using a congruent copy.
This relationship immediately leads to another crucial theorem concerning triangles: Triangles on the same base (or on equal bases) and lying between the same parallel lines are equal in area. Since each triangle is half the area of a parallelogram on the same base and between the same parallels (and those parallelograms would have equal area), the triangles must also have equal areas. The converse of this theorem is also significant and often explored: Triangles on the same base (or equal bases) and possessing equal areas must necessarily lie between the same parallel lines.
Another important consequence derived and utilized is the property of a triangle's median (a line segment joining a vertex to the midpoint of the opposite side). It is proven that a median of a triangle divides it into two triangles of equal areas. This property is frequently used in problems involving area division and comparison within triangles.
Throughout this chapter, the emphasis is less on formulaic calculation ($Area = base \times height$ or $Area = \frac{1}{2} \times base \times height$) and more on applying these fundamental theorems to compare areas directly based on the geometric setup presented in diagrams. Students will practice using these theorems to solve problems that require proving specific area relationships between different parts of a figure or calculating areas indirectly by relating them to other figures with known area relationships. It's a chapter focused on developing strong geometric reasoning about spatial extent.
Figures on Same Base and Between the Same Parallels
In this chapter, we will delve into the concept of the area of plane figures, with a particular focus on parallelograms and triangles.
We will explore important theorems that relate the areas of figures sharing specific geometric properties, such as having the same base and lying between the same parallel lines. These principles are fundamental tools for solving a wide range of problems involving area.
Understanding Key Concepts
Area of a Plane Figure
The area of a closed plane figure is the measure of the two-dimensional space or region enclosed by its boundary. It quantifies the amount of surface the figure covers.
Area is always a non-negative value and is measured in square units, such as square centimetres ($cm^2$), square metres ($m^2$), etc.
A crucial property is that congruent figures have equal areas. If two figures are congruent, they can perfectly overlap, meaning they must enclose the same area.
However, the converse of this statement is not always true: two figures with equal areas are not necessarily congruent. For example, a rectangle of $6 \text{ cm} \times 4 \text{ cm}$ has an area of $24 \text{ cm}^2$. A square with a side length of $\sqrt{24} \text{ cm}$ also has an area of $24 \text{ cm}^2$. Despite having the same area, the rectangle and the square are not congruent because their shapes are different.
Figures on the Same Base
Two geometric figures are said to be on the same base if they share a common side. This common side is referred to as their base.
Examples:
- In the figure below, parallelogram ABCD and $\triangle \text{PCD}$ share the common side DC. Therefore, they are on the same base DC.
- Here, trapezium ABCD and $\triangle \text{PCD}$ are on the same base DC.
Figures Between the Same Parallels
Two figures are said to lie between the same parallels if they share a common base (or equal bases) and the vertices opposite to the common base of each figure lie on a line parallel to the base.
This means the height of both figures, with respect to the common base, is the same because the distance between two parallel lines is always constant.
Examples:
- Parallelogram ABCD and $\triangle \text{PCD}$ are on the same base DC and between the same parallels AP and DC. The vertices A and B of the parallelogram and the vertex P of the triangle lie on the line AP, which is parallel to the base DC.
- $\triangle \text{ABC}$ and $\triangle \text{DBC}$ are on the same base BC and between the same parallels AD and BC. The vertices A and D lie on the line AD, which is parallel to the base BC.
Theorem 9.1: Parallelograms on the Same Base and Between the Same Parallels
Theorem 9.1. Parallelograms on the same base and between the same parallels are equal in area.
Proof:
Given:
Two parallelograms, ABCD and EFCD, which are on the same base DC and lie between the same parallel lines AF and DC.
To Prove:
Area(ABCD) = Area(EFCD).
Proof:
Our strategy is to prove that the two triangles at the ends, $\triangle \text{ADE}$ and $\triangle \text{BCF}$, are congruent.
In $\triangle \text{ADE}$ and $\triangle \text{BCF}$, we have:
$\angle \text{DAE} = \angle \text{CBF}$
[Corresponding angles, as AD $||$ BC and AF is the transversal] ... (i)
$\angle \text{AED} = \angle \text{BFC}$
[Corresponding angles, as ED $||$ FC and AF is the transversal] ... (ii)
Therefore, the third angles must also be equal:
$\angle \text{ADE} = \angle \text{BCF}$
[Angle sum property of a triangle]
Also, since ABCD is a parallelogram:
AD = BC
[Opposite sides of a parallelogram] ... (iii)
Using the Angle-Side-Angle (ASA) congruence rule with equations (i), (iii), and the conclusion from the angle sum property ($\angle \text{ADE} = \angle \text{BCF}$), we can state:
$\triangle \text{ADE} \cong \triangle \text{BCF}$ (By ASA congruence rule).
Since congruent figures have equal areas, we have:
Area($\triangle \text{ADE}$) = Area($\triangle \text{BCF}$)
... (iv)
Now, let's consider the areas of the parallelograms.
Area(ABCD) can be expressed as the sum of the areas of $\triangle \text{ADE}$ and trapezium EBCD.
Area(ABCD) = Area($\triangle \text{ADE}$) + Area(EBCD)
... (v)
Similarly, Area(EFCD) can be expressed as the sum of the areas of $\triangle \text{BCF}$ and trapezium EBCD.
Area(EFCD) = Area($\triangle \text{BCF}$) + Area(EBCD)
... (vi)
From equation (iv), we know that Area($\triangle \text{ADE}$) = Area($\triangle \text{BCF}$). Substituting this into equation (vi):
Area(EFCD) = Area($\triangle \text{ADE}$) + Area(EBCD)
Comparing this with equation (v), we get:
Area(ABCD) = Area(EFCD)
Hence, the theorem is proved.
Alternate Proof (Using Area Formula)
The area of any parallelogram is given by the formula: Area = base $\times$ height.
For parallelogram ABCD, the base is DC and the height is the perpendicular distance between the parallel lines AF and DC. Let this height be $h$.
Area(ABCD) = DC $\times$ $h$.
For parallelogram EFCD, the base is also DC, and since it lies between the same parallel lines AF and DC, its height is also $h$.
Area(EFCD) = DC $\times$ $h$.
From these two expressions, it is clear that:
Area(ABCD) = Area(EFCD)
Theorem 9.2: Area of a Triangle relative to a Parallelogram
Theorem 9.2. If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Proof:
Given:
A parallelogram ABCD and a triangle $\triangle \text{PBC}$ are on the same base BC and between the same parallel lines AP and BC.
To Prove:
Area($\triangle \text{PBC}$) = $\frac{1}{2}$ Area(ABCD).
Construction:
Through point C, draw a line CQ parallel to BP, such that it intersects the line AP at point Q.
Proof:
First, consider the quadrilateral PBCQ.
PQ $\parallel$ BC
(Given that A, P are on a line parallel to BC)
BP $\parallel$ CQ
(By construction)
Since both pairs of opposite sides are parallel, PBCQ is a parallelogram.
Now, consider the two parallelograms, ABCD and PBCQ. They are on the same base BC and are between the same parallels AQ and BC.
According to Theorem 9.1 (Parallelograms on the same base and between the same parallels are equal in area), we have:
Area(ABCD) = Area(PBCQ)
... (i)
Next, consider the parallelogram PBCQ. The line segment PC is a diagonal. We know that a diagonal of a parallelogram divides it into two congruent triangles.
Therefore, $\triangle \text{PBC} \cong \triangle \text{CQP}$.
Since congruent figures have equal areas:
Area($\triangle \text{PBC}$) = Area($\triangle \text{CQP}$)
Also, the area of the parallelogram is the sum of the areas of these two triangles:
Area(PBCQ) = Area($\triangle \text{PBC}$) + Area($\triangle \text{CQP}$) = Area($\triangle \text{PBC}$) + Area($\triangle \text{PBC}$)
Area(PBCQ) = 2 $\times$ Area($\triangle \text{PBC}$)
This implies:
Area($\triangle \text{PBC}$) = $\frac{1}{2}$ Area(PBCQ)
... (ii)
From equations (i) and (ii), by substituting Area(PBCQ) with Area(ABCD) in equation (ii), we get:
Area($\triangle \text{PBC}$) = $\frac{1}{2}$ Area(ABCD)
Hence, the theorem is proved.
Alternate Proof (Using Area Formula)
Let the common base be $b$ (length of BC) and the common height be $h$ (perpendicular distance between AP and BC).
Area of $\triangle \text{PBC} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times h$.
Area of parallelogram ABCD = $\text{base} \times \text{height} = b \times h$.
Comparing the two formulas, it is evident that Area($\triangle \text{PBC}$) is half of Area(ABCD).
Theorem 9.3: Triangles on the Same Base and Between the Same Parallels
Theorem 9.3. Triangles on the same base (or equal bases) and between the same parallels are equal in area.
Proof:
Given:
Two triangles $\triangle \text{ABC}$ and $\triangle \text{PBC}$ are on the same base BC and between the same parallels AP and BC.
To Prove:
Area($\triangle \text{ABC}$) = Area($\triangle \text{PBC}$).
Construction:
Through B, draw a line BD parallel to AC, which meets the line AP at D. This forms parallelogram DABC.
Proof:
We have constructed a parallelogram DABC which is on the base BC and between the parallels AP and BC.
From Theorem 9.2, we know that the area of a triangle is half the area of a parallelogram on the same base and between the same parallels.
For $\triangle \text{ABC}$ and parallelogram DABC:
Area($\triangle \text{ABC}$) = $\frac{1}{2}$ Area(DABC)
... (i)
Similarly, for $\triangle \text{PBC}$ and parallelogram DABC:
Area($\triangle \text{PBC}$) = $\frac{1}{2}$ Area(DABC)
... (ii)
From equations (i) and (ii), we can conclude that:
Area($\triangle \text{ABC}$) = Area($\triangle \text{PBC}$)
Hence, the theorem is proved.
Alternate Proof (Using Area Formula)
Let the common base be $b$ (length of BC) and the common height be $h$ (perpendicular distance between AP and BC).
Area($\triangle \text{ABC}$) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times h$.
Area($\triangle \text{PBC}$) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times h$.
Therefore, Area($\triangle \text{ABC}$) = Area($\triangle \text{PBC}$).
Converse of Theorem 9.3
Theorem 9.4. Triangles on the same base (or equal bases) having equal areas lie between the same parallels.
Proof:
Given:
Two triangles $\triangle \text{ABC}$ and $\triangle \text{PBC}$ are on the same base BC and Area($\triangle \text{ABC}$) = Area($\triangle \text{PBC}$).
To Prove:
The line segment AP is parallel to BC (i.e., the triangles lie between the same parallels).
Construction:
Draw the altitude from A to BC, meeting BC at D. Let AD = $h_1$.
Draw the altitude from P to BC, meeting BC at E. Let PE = $h_2$.
Proof:
The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Area($\triangle \text{ABC}$) $= \frac{1}{2} \times \text{BC} \times h_1$
... (i)
Area($\triangle \text{PBC}$) $= \frac{1}{2} \times \text{BC} \times h_2$
... (ii)
We are given that the areas are equal.
Area($\triangle \text{ABC}$) = Area($\triangle \text{PBC}$)
(Given)
Using equations (i) and (ii), we can write:
$\frac{1}{2} \times \text{BC} \times h_1 = \frac{1}{2} \times \text{BC} \times h_2$
Since the base BC has a non-zero length, we can cancel $\frac{1}{2} \times \text{BC}$ from both sides.
$h_1 = h_2$
This result means that the heights of both triangles with respect to the common base BC are equal. In other words, the perpendicular distance from point A to the line BC is the same as the perpendicular distance from point P to the line BC.
Two points that are on the same side of a line and are equidistant from it must lie on a line parallel to the given line.
Therefore, the line passing through A and P must be parallel to the line BC.
AP $\parallel$ BC
Hence, the theorem is proved.
Example 1. ABCD is a parallelogram and AE $\perp$ DC, CF $\perp$ AD. If AB = 10 cm, AE = 8 cm, and CF = 12 cm, find AD.
Answer:
Given:
Parallelogram ABCD. AB = 10 cm, AE = 8 cm, CF = 12 cm. AE is perpendicular to DC ($\text{AE} \perp \text{DC}$) and CF is perpendicular to AD ($\text{CF} \perp \text{AD}$).
To Find:
The length of side AD.
Solution:
In a parallelogram, opposite sides are equal in length.
$\text{DC} = \text{AB}$
(Opposite sides of parallelogram)
Given AB = 10 cm, so DC = 10 cm.
$\text{DC} = 10 \text{ cm}$
The area of a parallelogram can be calculated using the formula: Area = base $\times$ corresponding height.
Using base DC and its corresponding altitude AE:
Area(ABCD) = $\text{DC} \times \text{AE}$
Substitute the given values:
Area(ABCD) = $10 \text{ cm} \times 8 \text{ cm}$
Area(ABCD) = $80 \text{ cm}^2$
... (1)
Now, let's use side AD as the base and its corresponding altitude CF. The area of the parallelogram remains the same regardless of the base chosen.
Area(ABCD) = $\text{AD} \times \text{CF}$
We know Area(ABCD) = $80 \text{ cm}^2$ from equation (1) and the length of CF = 12 cm. Let the length of AD be $x$ cm.
$80 = x \times 12$
To find $x$ (which is AD), divide both sides by 12:
$x = \frac{80}{12}$
Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, 4:
$x = \frac{\cancel{80}^{20}}{\cancel{12}_{3}}$
$x = \frac{20}{3}$
The length of AD is $\frac{20}{3}$ cm.
If needed, this can be expressed as a mixed number ($6 \frac{2}{3}$ cm) or a decimal approximation (approximately 6.67 cm).
Example 2. In $\triangle \text{ABC}$, E is the midpoint of median AD. Show that Area($\triangle \text{BED}$) = $\frac{1}{4}$ Area($\triangle \text{ABC}$).
Answer:
Given:
In $\triangle \text{ABC}$, AD is the median to side BC.
E is the midpoint of the median AD.
To Prove:
Area($\triangle \text{BED}$) = $\frac{1}{4}$ Area($\triangle \text{ABC}$).
Proof:
We will use the property that a median of a triangle divides it into two triangles of equal areas.
In $\triangle \text{ABC}$, AD is the median. Therefore, it divides $\triangle \text{ABC}$ into two triangles of equal area, $\triangle \text{ABD}$ and $\triangle \text{ACD}$.
Area($\triangle \text{ABD}$) = Area($\triangle \text{ACD}$)
(Median divides a triangle into two triangles of equal areas)
This means that the area of $\triangle \text{ABD}$ is half the area of $\triangle \text{ABC}$.
Area($\triangle \text{ABD}$) = $\frac{1}{2}$ Area($\triangle \text{ABC}$)
... (1)
Now, consider $\triangle \text{ABD}$. We are given that E is the midpoint of the side AD. Therefore, BE is the median of $\triangle \text{ABD}$.
The median BE divides $\triangle \text{ABD}$ into two triangles of equal area, $\triangle \text{ABE}$ and $\triangle \text{BED}$.
Area($\triangle \text{BED}$) = Area($\triangle \text{ABE}$)
This means that the area of $\triangle \text{BED}$ is half the area of $\triangle \text{ABD}$.
Area($\triangle \text{BED}$) = $\frac{1}{2}$ Area($\triangle \text{ABD}$)
... (2)
Now, substitute the value of Area($\triangle \text{ABD}$) from equation (1) into equation (2):
Area($\triangle \text{BED}$) = $\frac{1}{2} \left( \frac{1}{2} \text{ Area}(\triangle\text{ABC}) \right)$
Area($\triangle \text{BED}$) = $\frac{1}{4}$ Area($\triangle \text{ABC}$)
Hence, proved.
Alternate Proof:
We will use the property that a median divides a triangle into two triangles of equal area.
Step 1: In $\triangle \text{ABC}$, AD is the median to BC.
Area($\triangle \text{ABD}$) = Area($\triangle \text{ACD}$)
... (i)
Step 2: In $\triangle \text{EBC}$, D is the midpoint of BC, so ED is the median.
Area($\triangle \text{EBD}$) = Area($\triangle \text{ECD}$)
... (ii)
Step 3: Subtracting equation (ii) from equation (i):
Area($\triangle \text{ABD}$) - Area($\triangle \text{EBD}$) = Area($\triangle \text{ACD}$) - Area($\triangle \text{ECD}$)
From the figure, we can see that:
Area($\triangle \text{ABD}$) - Area($\triangle \text{EBD}$) = Area($\triangle \text{ABE}$)
Area($\triangle \text{ACD}$) - Area($\triangle \text{ECD}$) = Area($\triangle \text{ACE}$)
Therefore,
Area($\triangle \text{ABE}$) = Area($\triangle \text{ACE}$)
... (iii)
Step 4: Now consider $\triangle \text{ABD}$. E is the midpoint of AD, so BE is the median.
Area($\triangle \text{BED}$) = Area($\triangle \text{ABE}$)
... (iv)
Step 5: From equations (ii), (iii) and (iv), we can relate the areas of all four small triangles:
Area($\triangle \text{BED}$) = Area($\triangle \text{ABE}$) [From (iv)]
Area($\triangle \text{ABE}$) = Area($\triangle \text{ACE}$) [From (iii)]
Area($\triangle \text{ACE}$) = Area($\triangle \text{ACD}$) - Area($\triangle \text{ECD}$) = Area($\triangle \text{ABD}$) - Area($\triangle \text{EBD}$)
Combining them, we find:
Area($\triangle \text{ABE}$) = Area($\triangle \text{BED}$) = Area($\triangle \text{ACE}$) = Area($\triangle \text{CED}$)
Let Area($\triangle \text{BED}$) = $x$. Then,
Area($\triangle \text{ABC}$) = Area($\triangle \text{ABE}$) + Area($\triangle \text{BED}$) + Area($\triangle \text{ACE}$) + Area($\triangle \text{CED}$)
Area($\triangle \text{ABC}$) = $x + x + x + x = 4x$
Area($\triangle \text{ABC}$) = 4 $\times$ Area($\triangle \text{BED}$)
This gives:
Area($\triangle \text{BED}$) = $\frac{1}{4}$ Area($\triangle \text{ABC}$)
Hence, proved.
Area of Polygonal Regions
The area of a polygonal region is the measure of the surface enclosed by the boundary of the polygon. Area is a fundamental concept in geometry and is measured in square units (e.g., square centimetres ($\text{cm}^2$), square metres ($\text{m}^2$), etc.).
Area Axioms
The concept of area for polygonal regions is built upon a set of fundamental assumptions, or axioms. These are accepted without proof and form the basis for deriving all area formulas.
- Existence and Positivity Axiom: Every polygonal region has a unique area, which is a positive real number.
- Congruence Axiom: If two polygonal regions are congruent, they have equal areas. (This was used in proving Theorem 9.1).
- Area Addition Axiom: If a polygonal region R is the union of two non-overlapping polygonal regions R1 and R2, then the area of the entire region is the sum of the areas of the individual regions. That is, Area(R) = Area(R1) + Area(R2).
- Unit Axiom: The area of a square with a side length of 1 unit is defined as 1 square unit. This extends to defining the area of a rectangle with length $l$ and breadth $b$ as their product, $l \times b$. This serves as the starting point for calculating the areas of other figures.
Formulas for Areas of Common Polygons
Using the axioms and theorems discussed, we can establish the standard formulas for the areas of various polygons.
1. Area of a Rectangle
From the Unit Axiom, the area of a rectangle with length $l$ and breadth $b$ is given by:
Area = $l \times b$
2. Area of a Square
A square is a special rectangle where length and breadth are equal. If the side length is $s$, then $l=s$ and $b=s$.
Area = $s \times s = s^2$
3. Area of a Parallelogram
The area of a parallelogram with base $b$ and corresponding height $h$ is equal to the area of a rectangle with the same base and height.
Area = $b \times h$
4. Area of a Triangle
From Theorem 9.2, we know that the area of a triangle is half the area of a parallelogram on the same base and between the same parallels. A parallelogram with base $b$ and height $h$ has an area of $b \times h$.
Therefore, the area of a triangle with base $b$ and corresponding height $h$ is:
Area = $\frac{1}{2} \times b \times h$
5. Area of a Rhombus
A rhombus can be divided by its diagonals, $d_1$ and $d_2$, into four congruent right-angled triangles. Alternatively, we can view it as two congruent triangles with base $d_1$ and height $\frac{d_2}{2}$.
Derivation:
Area(Rhombus ABCD) = Area($\triangle \text{ABC}$) + Area($\triangle \text{ADC}$)
Let AC = $d_1$ and BD = $d_2$. Let the diagonals intersect at O.
Area($\triangle \text{ABC}$) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{AC} \times \text{BO} \ $$ = \frac{1}{2} \times d_1 \times \frac{d_2}{2} = \frac{1}{4} d_1 d_2$
Area($\triangle \text{ADC}$) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{AC} \times \text{DO} \ $$ = \frac{1}{2} \times d_1 \times \frac{d_2}{2} = \frac{1}{4} d_1 d_2$
Total Area = $\frac{1}{4} d_1 d_2 + \frac{1}{4} d_1 d_2 = \frac{2}{4} d_1 d_2 = \frac{1}{2} d_1 d_2$
Thus, the area of a rhombus is half the product of its diagonals.
Area = $\frac{1}{2} d_1 d_2$
6. Area of a Trapezium
A trapezium with parallel sides $a$ and $b$ and height $h$ can be divided into two triangles by drawing a diagonal.
Derivation:
Let the trapezium be ABCD with AB $\parallel$ DC. Let AB = $a$ and DC = $b$. Draw diagonal AC.
Area(Trapezium ABCD) = Area($\triangle \text{ABC}$) + Area($\triangle \text{ADC}$)
Area($\triangle \text{ABC}$) = $\frac{1}{2} \times \text{base AB} \times \text{height} = \frac{1}{2} \times a \times h$
Area($\triangle \text{ADC}$) = $\frac{1}{2} \times \text{base DC} \times \text{height} = \frac{1}{2} \times b \times h$
Total Area = $\frac{1}{2} a h + \frac{1}{2} b h = \frac{1}{2} h (a + b)$
Thus, the area of a trapezium is half the product of the height and the sum of its parallel sides.
Area = $\frac{1}{2} (a + b) h$
Theorem 9.5: Median of a Triangle Divides it into Equal Areas
Theorem 9.5. A median of a triangle divides it into two triangles of equal areas.
Proof:
Given:
A triangle $\triangle \text{ABC}$. AD is a median to side BC, which means D is the midpoint of BC.
To Prove:
Area($\triangle \text{ABD}$) = Area($\triangle \text{ACD}$).
Construction:
Draw an altitude AM from vertex A to the line containing BC, such that AM $\perp$ BC. Point M lies on the line containing BC.
Proof:
Consider $\triangle \text{ABD}$. The base is BD and the corresponding height is AM (the perpendicular distance from vertex A to the line containing the base BD, which is the line BC).
Area($\triangle \text{ABD}$) $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{BD} \times \text{AM}$
... (1)
Consider $\triangle \text{ACD}$. The base is CD and the corresponding height is AM (the perpendicular distance from vertex A to the line containing the base CD, which is the line BC).
Area($\triangle \text{ACD}$) $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{CD} \times \text{AM}$
... (2)
Since AD is the median to side BC, D is the midpoint of BC. By the definition of a midpoint,
$\text{BD} = \text{CD}$
(D is the midpoint of BC) ... (3)
Substitute the equality from equation (3) into equation (1):
Area($\triangle \text{ABD}$) $= \frac{1}{2} \times \text{CD} \times \text{AM}$
Now, compare this expression with equation (2). We see that both expressions are identical.
Therefore, Area($\triangle \text{ABD}$) = Area($\triangle \text{ACD}$).
Thus, a median of a triangle divides it into two triangles of equal areas. This completes the proof.