256. The Median's Equal Share · splitting any triangle's area perfectly in two

Try to break it

Drag A, B, or C. D stays at the midpoint of BC, so the two sub-triangles △ABD and △ADC share the same base length (BD = DC) and the same height (the perpendicular from A to BC). Their areas are always equal. Try to drag the vertices so the two sub-areas drift apart — impossible.

How you build it

Construct a triangle with median and altitude.

• Point tool: mark point A, the first vertex.
• Point tool: mark point B, the second vertex.
• Point tool: mark point C, the third vertex.
• Segment tool: join A to B.
• Segment tool: join B to C — the base.
• Segment tool: join C to A — triangle ABC is complete.
• Midpoint tool: click B then C — D drops exactly at the midpoint of BC, so BD = DC.
• Segment tool: join A to D — the median.
• Perp tool: from A, drop a perpendicular to BC — its length is the common height of ΔABD and ΔADC. Equal bases (BD = DC) and equal height means equal area.

The proof, step by step

Prove that a median divides a triangle into two triangles of equal area.

• AD is the median of ΔABC, so BD = DC.
• AP is the altitude to BC, so AP ⊥ BC.
• Area(ΔABD) = 1/2 × BD × AP.
• Area(ΔADC) = 1/2 × DC × AP.
• Since BD = DC, Area(ΔABD) = Area(ΔADC).
• Each area is exactly half of Area(ΔABC).

Worked example

In ΔABC, AD is a median to side BC. If the area of ΔABC is 60 cm², find the area of ΔABD.

Since AD is a median of ΔABC, it divides the triangle into two triangles of equal area. Therefore, Area(ΔABD) = 1/2 × Area(ΔABC) = 1/2 × 60 = 30 cm².

• 15 cm²
• 30 cm² — correct
• 40 cm²
• 60 cm²