Triangles between parallel lines

Imagine a triangle with one fixed base, and its third vertex sliding along a line parallel to that base. As the vertex moves, the triangle changes shape , it becomes tall and thin, or wide and squat. But here is a small marvel: its area stays exactly the same. This is one of the most quietly beautiful facts in plane geometry, and it has many uses.

Concept

The set-up. Draw a horizontal base $BC$. Draw a line $\ell$ parallel to $BC$, somewhere above. Choose any point $A$ on $\ell$, and form $\triangle ABC$. Now slide $A$ along $\ell$ to get $A'$, $A''$, etc. , different triangles, same base.

The claim. All such triangles have the same area.

Why. The area formula is $\dfrac{1}{2} \times \text{base} \times \text{height}$. The base $BC$ doesn't change. The height , the perpendicular distance from $A$ to the line through $BC$ , is the same distance between the two parallel lines regardless of where $A$ sits on $\ell$. So both factors stay constant, and so does the area.

A neat consequence. If two triangles share a common base and their third vertices lie on a common line parallel to that base, they have equal areas , even though they may look very different.

The shortest-perimeter triangle. Among all the triangles in the family, they all have the same area, but they do not have the same perimeter. The perimeter changes: a thin slanting triangle has a long side, while the symmetric (isoceles) one has the shortest possible total length for the other two sides. In fact:

• The triangle with $A$ at the foot of the perpendicular bisector of $BC$ on $\ell$ is the isosceles triangle in the family , and it has the minimum perimeter (because by the reflection / mirror argument, this is the shortest path from $B$ to $C$ via the line $\ell$).
• There is no maximum perimeter , you can make the triangle as long and stretched as you like.

Why care? This idea pops up everywhere:

• It lets you compare or equate areas of seemingly different triangles by checking a single "same base, same parallel" fact.
• In medians: the median from a vertex divides the opposite side in half, so the two resulting triangles share the same base length and have the same vertex (same height). Hence they have equal areas. This is why a median splits a triangle into two equal-area pieces.
• In rectangles and parallelograms: a diagonal splits them into two equal-area triangles , same base, same height.

Worked examples

Example 1. Triangles $\triangle ABC$ and $\triangle A'BC$ share base $BC = 6$ cm. The third vertices $A$ and $A'$ lie on a line $\ell$ parallel to $BC$, with $\ell$ at distance $4$ cm from $BC$. Find their areas.

• Both areas $= \dfrac{1}{2} \times 6 \times 4 = 12 \mathrm{cm}^2$.

Example 2. $ABCD$ is a rectangle with $AB = 8$ cm, $BC = 5$ cm. Diagonal $AC$ divides it into $\triangle ABC$ and $\triangle ACD$. Find their areas.

• Each $= \dfrac{1}{2} \times 8 \times 5 = 20 \mathrm{cm}^2$. (Each is half the rectangle.)

Example 3. In $\triangle ABC$, median $AM$ is drawn from $A$ to the midpoint $M$ of $BC$. If the area of $\triangle ABC$ is $40 \mathrm{cm}^2$, find the area of $\triangle ABM$.

• The median divides $\triangle ABC$ into two equal-area triangles. So $\text{area}(\triangle ABM) = 20 \mathrm{cm}^2$.

Example 4. Two triangles, $\triangle XDC$ and $\triangle YDC$, share base $DC$ on the bottom side of a rectangle, with $X$ and $Y$ on the top side of the same rectangle. Are their areas equal?

• Yes. They share the base $DC$, and both $X$ and $Y$ lie on the same line parallel to $DC$. Same base, same height , same area.

Try it yourself

1. Triangles share base $BC = 10$ cm with their third vertices on a line $4$ cm from $BC$. Find each area.
2. A rectangle of $12 \times 5$ has a diagonal drawn. Each triangle has area?
3. Median $AM$ of $\triangle ABC$ has area $\triangle ABC = 30$. Area of $\triangle ABM$?
4. Why do all triangles between the same two parallel lines on the same base have equal area?
5. Among such triangles, which has the smallest perimeter?
6. Two triangles $\triangle ABC$ and $\triangle DEF$ have bases $BC = EF = 6$ cm and heights both $4$ cm. Are their areas equal?
7. A rectangle has both diagonals drawn, creating four triangles. Show all four have equal area.
8. Triangle with vertices $A(0, 4), B(0, 0), C(6, 0)$. Triangle with vertices $A'(3, 4), B(0, 0), C(6, 0)$. Same area? Why?

Activity

Sliding vertex. On a piece of paper, draw a horizontal segment $BC$. Above it, draw a horizontal line $\ell$. Take a pin and a thread loop. Place the pin at any point on $\ell$, stretch the thread to $B$ and to $C$ to mark a triangle. Repeat with the pin at three more points on $\ell$, each time drawing the triangle with thread. The triangles will look different , yet by measuring (or by the formula), they all enclose the same area. The pin slides on $\ell$, but the triangle's area is "stuck" at $\dfrac{1}{2} \times BC \times \text{distance between lines}$.