The relation of triangles standing on the same base and lying between the same parallels

Picture two parallel lines drawn across a page. Mark a segment AB on the lower line. Choose any point on the upper line and join it to A and to B: you get a triangle standing on the base AB. Choose a different point on the upper line and you get a second triangle on the very same base. These two triangles stand on the same base AB and lie between the same parallel lines.

The height of a triangle is the perpendicular distance from its apex to the line of its base. Two parallel lines stay the same distance apart all along their length, so wherever the apex sits on the upper line, the height is the same. A triangle's area is \( \frac{1}{2} \times \text{base} \times \text{height} \), so the same base with the same height must give the same area.

Three facts about areas are already known, and every proof in this topic leans on them:

• Parallelograms standing on the same base and lying between the same parallels are equal in area.
• When a triangle and a parallelogram stand on the same base and lie between the same parallels, the area of the triangle is half the area of the parallelogram.
• A diagonal divides a parallelogram into two triangles of equal area.

In the figure below, AB = 8 cm lies on the lower line and EC is parallel to AB. AE is perpendicular to AB and AE = 6 cm. Triangles ABE and ABC both stand on AB, and their apexes E and C both lie on the line EC. In the working, \( \triangle ABE \) written in an equation means the area of triangle ABE.

Step 1: Triangle ABE has a right angle at A, so its base is AB = 8 cm and its height is AE = 6 cm:

\[ \triangle ABE = \frac{1}{2} \times 8 \times 6 = 24 \text{ cm}^2 \]

Step 2: Triangle ABC also has base AB = 8 cm. Its height is the distance from C down to AB, and since EC is parallel to AB this distance equals AE = 6 cm:

\[ \triangle ABC = \frac{1}{2} \times 8 \times 6 = 24 \text{ cm}^2 \]

Step 3: Comparing the two areas:

\[ \triangle ABE = \triangle ABC \]

Measuring worked for one pair of numbers. A proof shows the result is true for every such pair of triangles, whatever their size or shape. The proof is set out in the usual way:

• Given: Triangles ABE and ABC stand on the same base AB and lie between the parallel lines AB and EC.
• To prove: Area of \( \triangle ABE \) = area of \( \triangle ABC \).
• Construction: Draw AD parallel to BC, meeting EC at D. Now ABCD is a parallelogram.

The construction is the clever part. The two triangles cannot be compared with each other directly, but each one can be compared with the same parallelogram ABCD. Each triangle is exactly half of that parallelogram, and two things that are each half of the same thing are equal to each other. Proved.

Most proofs about area are short chains of statements. Each link is backed by one of the three area facts, by Theorem 3, or by one of the reasons below. Knowing them by name makes the reasons column easy to write.

One more fact is useful when a midpoint appears. A median joins a vertex to the midpoint of the opposite side. It cuts the triangle into two triangles with equal bases and the same height, so the two halves have equal area.

ABCD is a parallelogram and P is any point inside it. Prove that area of \( \triangle APB \) + area of \( \triangle PCD = \frac{1}{2} \) area of parallelogram ABCD.

• Given: ABCD is a parallelogram and P is a point inside it.
• To prove: \( \triangle APB + \triangle PCD = \frac{1}{2} \) parallelogram ABCD.
• Construction: Through P draw MN parallel to CD, with M on AD and N on BC. Then ABNM and MNCD are parallelograms.

Notice that P could be anywhere inside the parallelogram. The two remaining triangles, APD and BPC, must then make up the other half. A single line through P is enough to split the problem into two easy pieces. Proved.

PQRS is a parallelogram. X and Y are any two points on PS and SR respectively. Prove that area of \( \triangle PQY \) = area of \( \triangle QRX \).

• Given: PQRS is a parallelogram; X is on PS and Y is on SR.
• To prove: \( \triangle PQY = \triangle QRX \).

The two triangles do not share a base at all, so Theorem 3 cannot be used on them directly. Instead each one is measured against the whole parallelogram, using a different pair of its parallel sides each time. Proved.

PQRS is a parallelogram and O is any point on the side QR. SO is joined and produced, and PQ is produced, so that they meet at T. Prove that area of \( \triangle PQO \) = area of \( \triangle RTO \).

• Given: O is on side QR of parallelogram PQRS; PQ and SO produced meet at T.
• To prove: \( \triangle PQO = \triangle RTO \).
• Construction: Draw the diagonal PR.

Here two big triangles, PQR and TRS, are shown equal first. Then two equal pieces, POR and SOR, are cut away from them, and what is left over must be equal. When the triangles you want are the leftovers of bigger equal triangles, subtract. Proved.

ABCD is a parallelogram. M and N are the midpoints of BC and AD respectively. Prove that MN divides parallelogram ABCD into two parallelograms of equal area.

• Given: ABCD is a parallelogram; M and N are the midpoints of BC and AD.
• To prove: area of parallelogram ABMN = area of parallelogram CDNM.

In triangle PQR, M is a point on PQ and N is a point on PR such that MN // QR. Prove that area of \( \triangle PQN \) = area of \( \triangle PRM \).

This is the opposite move to the last example. The equal triangles MNQ and MNR are small, and each becomes one of the wanted triangles once the top triangle PMN is stuck onto it. When the triangles you want are built from equal pieces plus a common piece, add. Proved.

Sometimes the areas are known to be equal and a pair of parallel lines has to be proved. Suppose triangles ABE and ABF stand on the same base AB, on the same side of it, and have equal areas. Equal areas on equal bases mean equal heights, so E and F are the same distance from AB. The line EF therefore runs parallel to AB.

Before writing the table, spend a minute reading the figure. These questions, in this order, find the route in almost every problem:

• Mark every pair of parallel lines in the figure, including any that the construction will add.
• For each triangle in the question, find a side that lies on one of those lines and check whether its apex lies on the parallel line.
• If two triangles share that base, use Theorem 3. If a triangle shares its base with a parallelogram, it is half of that parallelogram.
• If the wanted triangles do not share a base, look for a third shape, often the whole parallelogram, that each can be compared with.
• Finally decide whether a common triangle must be added to both sides or taken away from both sides, and name the axiom you use.