Triangle Area and Side–Angle Theorems

Try this

Drag $D$ : the areas of $△ A B D$ and $△ A D C$ both change, but the split points of the “Area” bar and the “Base” bar always line up.
Drag $D$ close to the midpoint of $B C$ : the two areas become equal. A median cuts a triangle into two halves of equal area.
Drag the vertex $A$ : both areas grow or shrink together, but their ratio does not move at all, because it depends only on $B D : D C$ .
The dashed line $A H$ is the common height of both triangles. That is the whole secret: $S = \frac{1}{2} \times base \times height$ , so with equal heights the area follows the base.

Predict first

$△ A B C$ has area $30$ . $D$ lies on $B C$ with $B D : D C = 2 : 3$ . What is the area of $△ A B D$ ?

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

Why: the equal-height model

Area ratio = base ratio

Step 1 / 4

Two triangles
$A D$ cuts $△ A B C$ into $△ A B D$ (orange) and $△ A D C$ (blue).

$S_{△ A B C} = S_{△ A B D} + S_{△ A D C}$
Find the common height
Write down the areas
Divide: the height cancels

It also works the other way round: with the same base, area ratio = height ratio. With the same base and the same height, the areas are equal, even if the shapes look completely different.

A

can only slide along the dashed line parallel to

B C

. The shape changes. Does the area?

The common-side theorem

Two triangles $△ P A B$ and $△ Q A B$ share the side $A B$ . Join $P Q$ , and let line $P Q$ meet line $A B$ at $M$ . Then

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

Drag

P

and

Q

to the same side of

A B

, then to opposite sides. Do the splits of the two bars still line up?

Proof of the common-side theorem: build a bridge

Step 1 / 4

No common height
$△ P A B$ and $△ Q A B$ have the same base ( $A B$ ), so their area ratio is the ratio of their heights. But those two heights are awkward to use in this figure.
Look at it differently: use PM as a base
Do it again for B
Put them together

What makes the common-side theorem so useful: you never have to draw a height, and it does not care whether $P$ and $Q$ are on the same side or on opposite sides. As long as the figure has the intersection point, you can use it. Many “find the ratio of two segments” problems take one step once you turn them into area ratios.

Sides and angles: the bigger side faces the bigger angle

In any triangle, the longer side is opposite the larger angle, and the converse is true too. Drag the vertices below and check whether the longest side and the largest angle always face each other.

The thick line is the longest side; the filled sector is the largest angle.

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: The area of $△ A B C$ does not change

The base $B C$ stays put and the distance from $A$ to $B C$ stays the same, so the area does not change.

Practice

Exercise 1Basic

In $△ A B C$ , $D$ lies on $B C$ with $B D = 4$ and $D C = 7$ . The area of $△ A B D$ is $12$ . Find the area of $△ A D C$ .

Exercise 2Basic

$△ A B C$ has area $36$ . $D$ is the midpoint of $B C$ , and $E$ lies on $A D$ with $A E : E D = 2 : 1$ . Find the area of $△ B D E$ .

Exercise 3Advanced

The diagonals $A C$ and $B D$ of quadrilateral $A B C D$ meet at $O$ . We know $S_{△ A B C} = 15$ and $S_{△ A D C} = 10$ . Find $B O : O D$ (as a single number).

Exercise 4Challenge

$△ A B C$ has area $1$ . $D$ , $E$ , $F$ lie on $B C$ , $C A$ , $A B$ with $B D : D C = C E : E A = A F : F B = 1 : 2$ . The lines $A D$ , $B E$ , $C F$ cross each other and enclose a small triangle. Find its area.

Try this

Why: the equal-height model

Area ratio = base ratio

The common-side theorem

Proof of the common-side theorem: build a bridge

Sides and angles: the bigger side faces the bigger angle

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Comparing bases when the heights are different

Comparing the wrong segments in the common-side theorem