The Area Method

In one sentence

A triangle's area can be written in many ways: $\frac{1}{2} ah$ , $\frac{1}{2} ab sin C$ , $\frac{1}{2} ∣ x_{1} y_{2} - x_{2} y_{1} ∣$ … Write the same area in two ways and you get an equation. That is the area method.

The handiest tool is the common-side theorem: if line $P Q$ meets line $A B$ at $M$ , then

\frac{P M}{QM} = \frac{S _{△ P A B}}{S _{△ Q A B}} .

The two triangles share the base $A B$ , so their areas are in the ratio of their heights; and the heights are in the ratio $P M : QM$ (two similar right triangles). So segment ratio = area ratio, and you no longer need to hunt for similar triangles.

Predict first

In $△ A B C$ , $D$ lies on $B C$ with $B D : D C = 2 : 1$ , $E$ is the midpoint of $A D$ , and line $B E$ meets $A C$ at $F$ . What is $A F : F C$ ?

1 : 1

2 : 3

1 : 2

3 : 4

Choose an answer first; then you can see the result.

Example: BD : DC = 2 : 1, E is the midpoint of AD, BE meets AC at F; find AF : FC

Step 1 / 4

Find the common side
$F$ lies on line $A C$ and also on line $B E$ . Take $B E$ as the common side: the ratio of the areas of $△ A B E$ and $△ C B E$ is $A F : F C$ .

$\frac{A F}{F C} = \frac{S _{△ A B E}}{S _{△ C B E}}$
Compute S△ABE
Compute S△CBE
Conclusion

Stop 1 · Identities and Area in a Triangle (Vol. 3)

An angle bisector cuts a triangle into two pieces. Write their areas with $\frac{1}{2} ab sin C$ , and both the angle bisector theorem and the length of the bisector take one line each.

Example: in △ABC, AD bisects ∠A (D on BC), AB = c, AC = b

Step 1 / 4

Common height: ratio of bases
$△ A B D$ and $△ A C D$ have the same height (from $A$ to $B C$ ), so their areas are in the ratio $B D : D C$ .

$\frac{B D}{D C} = \frac{S _{△ A B D}}{S _{△ A C D}}$
Common angle: ratio of side products
Add the areas
The length of the bisector

Stop 2 · The Inner Product (Vol. 4)

Since $sin^{2} θ = 1 - cos^{2} θ$ , the area of the triangle spanned by two vectors can be written using only inner products; in coordinates it becomes the familiar determinant.

Example: the area of the triangle spanned by a = (3, 1) and b = (1, 2)

Step 1 / 4

The area formula
$S = \frac{1}{2} ∣ a ∣∣ b ∣ sin θ$ ; replace $sin θ$ using the inner product.

$S = \frac{1}{2} ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}$
Substitute
In coordinates
The two ways agree

Stop 3 · Basic Problems of the Coordinate Method (Vol. 7)

The distance from a point to a line is the height of a triangle: first compute the area from coordinates, then divide by the base.

One problem, several solutions

Given $A (1, 2)$ , $B (4, 6)$ , $C (5, 1)$ , find the distance from $C$ to line $A B$ .

The point-to-line distance formula

Line $A B$ has slope $\frac{4}{3}$ and equation $4 x - 3 y + 2 = 0$ .

$d = \frac{∣4 \times 5 - 3 \times 1 + 2∣}{4 ^{2} + 3 ^{2}} = \frac{19}{5}$ .

CostFind the equation of the line first, then use the formula.

$A B = (3, 4)$ , $A C = (4, - 1)$ , $S = \frac{1}{2} ∣3 \times (- 1) - 4 \times 4∣ = \frac{19}{2}$ .

$∣ A B ∣ = 5$ , so $d = \frac{2 S}{∣ A B ∣} = \frac{19}{5}$ .

CostNo equation of the line, just two vectors.

Practice

Exercise 1Basic

The legs of a right triangle are $6$ and $8$ . Find the altitude to the hypotenuse.

Exercise 2Basic

The medians $A M$ and $B N$ of $△ A B C$ meet at $G$ . Use the area method to find $A G : GM$ .

Exercise 3Advanced

In $△ A B C$ , $A B = 6$ , $A C = 3$ , $∠ A = 12 0^{\circ}$ , and the bisector $A D$ of $∠ A$ meets $B C$ at $D$ . Find $A D$ .

Exercise 4Advanced

Find the area of the triangle with vertices $O (0, 0)$ , $P (4, 1)$ , $Q (1, 3)$ .

The Area Method

In one sentence

Example: BD : DC = 2 : 1, E is the midpoint of AD, BE meets AC at F; find AF : FC

Stop 1 · Identities and Area in a Triangle (Vol. 3)

Example: in △ABC, AD bisects ∠A (D on BC), AB = c, AC = b

Stop 2 · The Inner Product (Vol. 4)

Example: the area of the triangle spanned by a = (3, 1) and b = (1, 2)

Stop 3 · Basic Problems of the Coordinate Method (Vol. 7)

The point-to-line distance formula

The area method

Practice

Mistake Museum · Exhibits from this stop

Different bases, yet the area ratio is taken as the ratio of heights

Forgetting the absolute value when finding an area from coordinates

Stops on this line