Area Chasing is the habit of comparing areas through shared bases, heights, ratios, and decompositions. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
$$[\triangle ABC] = \tfrac{1}{2}\,bh; \qquad \frac{[\triangle ABD]}{[\triangle ACD]} = \frac{BD}{CD} \text{ when } A,D \text{ share the line}$$
You compare regions by exploiting shared bases or heights so area ratios become simple length ratios.
Why It Works
1
The area of a triangle is $\tfrac12 \times \text{base} \times \text{height}$, so two triangles with the SAME height have areas in the ratio of their bases.
2
If $D$ is on segment $BC$, then $\triangle ABD$ and $\triangle ACD$ share the apex $A$ and have the same height from $A$ to line $BC$, giving $\dfrac{[\triangle ABD]}{[\triangle ACD]} = \dfrac{BD}{DC}$.
3
Likewise, triangles sharing a base have areas in the ratio of their heights.
4
Chaining these, you decompose a figure, write each piece's area as a ratio, and add or subtract to reach the target area.
Worked Examples
Example 1
In $\triangle ABC$ with area $48$, point $D$ lies on $BC$ with $BD:DC = 1:3$, and $E$ is the midpoint of $AD$. Find the area of $\triangle BEC$.
1
We want $[\triangle BEC]$, where $E$ lies on cevian $AD$. Use the common base $BC$ for both $\triangle ABC$ and $\triangle EBC$.
2
Triangles on the same base $BC$ have areas proportional to their heights, i.e. to the perpendicular distances of the apex ($A$ or $E$) from line $BC$.
3
Since $E$ is the midpoint of $AD$ and $D$ lies ON $BC$ (distance $0$), $E$'s distance to $BC$ is exactly half of $A$'s distance to $BC$.
4
Therefore $[\triangle BEC] = \tfrac12\,[\triangle ABC] = \tfrac12\cdot 48 = 24$. (The $BD:DC$ ratio is a distractor — it does not affect this height comparison.)
Answer:
$[\triangle BEC] = 24$
Warm-up
In $\triangle ABC$ with area $30$, point $D$ on $BC$ satisfies $BD:DC = 2:3$. Find the areas of $\triangle ABD$ and $\triangle ACD$.
1
$\triangle ABD$ and $\triangle ACD$ share the apex $A$ and the same height to line $BC$, so their areas are in the ratio $BD:DC = 2:3$.
2
Together they fill $\triangle ABC$ (area $30$), and $2 + 3 = 5$ parts.
3
So $[\triangle ABD] = \tfrac{2}{5}\cdot 30 = 12$ and $[\triangle ACD] = \tfrac{3}{5}\cdot 30 = 18$.
Answer:
$[\triangle ABD] = 12$, $[\triangle ACD] = 18$.
Contest level
In $\triangle ABC$, $D$ lies on $AB$ with $AD:DB = 1:2$ and $E$ lies on $AC$ with $AE:EC = 3:1$. If $[\triangle ABC] = 40$, find $[\triangle ADE]$.
1
$\triangle ADE$ shares angle $A$ with $\triangle ABC$, so $\dfrac{[\triangle ADE]}{[\triangle ABC]} = \dfrac{AD}{AB}\cdot\dfrac{AE}{AC}$ (the two-sides-times-sine area formula, common angle).
2
Here $\dfrac{AD}{AB} = \dfrac{1}{3}$ and $\dfrac{AE}{AC} = \dfrac{3}{4}$.
In $\triangle ABC$, points $D, E, F$ lie on $BC, CA, AB$ with $\dfrac{BD}{DC} = \dfrac{CE}{EA} = \dfrac{AF}{FB} = \dfrac{1}{2}$. Find the ratio $\dfrac{[\triangle DEF]}{[\triangle ABC]}$.
1
Cut off the three corner triangles. Consider $\triangle AFE$: it has $\dfrac{AF}{AB} = \dfrac{1}{3}$ and $\dfrac{AE}{AC} = \dfrac{EA}{AC} = \dfrac{2}{3}$ (since $CE:EA = 1:2$ gives $AE:AC = 2:3$).
2
So $\dfrac{[\triangle AFE]}{[\triangle ABC]} = \dfrac{AF}{AB}\cdot\dfrac{AE}{AC} = \dfrac{1}{3}\cdot\dfrac{2}{3} = \dfrac{2}{9}$. By the cyclic symmetry of the data, each corner triangle is also $\dfrac{2}{9}$.
3
The three corners together are $3\cdot\dfrac{2}{9} = \dfrac{6}{9} = \dfrac{2}{3}$ of $\triangle ABC$.
$\dfrac{[\triangle DEF]}{[\triangle ABC]} = \dfrac{1}{3}$ (subtract the three equal corner triangles).
Going Deeper
Generalization: for two triangles sharing an angle, $\dfrac{[\triangle]}{[\triangle']} = \dfrac{\text{product of the two enclosing sides}}{\text{product of the matching sides}}$, from $[\triangle] = \tfrac12 ab\sin\theta$. The example above subtracts the three corner triangles; the related Routh's theorem (for the triangle cut out by the three CEVIANS themselves, not the division points) gives $\dfrac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}$.
Where it appears: area ratios solve 'find the area of the inner region' problems on AMC/AIME without coordinates, and they prove the angle-bisector theorem, Ceva, and mass-point ratios. Whenever two regions share a base or an apex, an area ratio collapses to a length ratio.
Pitfall: the 'same height' argument requires the apex to lie on a line PARALLEL to the shared base (or the foot to lie on the base line). If you compare two triangles whose apexes are at different perpendicular distances, the area ratio is NOT the base ratio — confirm the heights are genuinely equal before equating.
Spot the Signal
Look for problems where the key step is comparing areas through shared bases, heights, ratios, and decompositions.
You can describe the hard part as comparing areas through shared bases, heights, ratios, and decompositions, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the diagram relation that calls for area chasing, then rewrite the givens around it.
Name the relation that makes Area Chasing legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Area Chasing just because the surface looks familiar; verify the required condition first.
Applying Area Chasing because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let area chasing reveal the geometric configuration; then compute only what remains.
Try it on:
Practice a contest problem where the key step is comparing areas through shared bases, heights, ratios, and decompositions.