255. Same Base, Same Height · why parallel lines lock in equal areas

Try to break it

Drag C and D along the top parallel line. △ABC and △ABD share base AB and share the same height (the perpendicular distance to the top line), so their areas stay exactly equal at every position. Try to drag C or D so one area pulls ahead of the other; impossible.

How you build it

Construct triangles ABC and ABD on the same base AB, between the same parallels.

• Point tool: mark A — the left end of the base.
• Point tool: mark B — the right end of the base.
• Segment tool: join A to B — the common base.
• Parallel tool: click a spot above, then click AB — both apexes will sit on this top parallel.
• Point tool: mark C on the top line — apex of the first triangle.
• Point tool: mark D elsewhere on the same top line — apex of the second triangle.
• Segment tool: join A to C.
• Segment tool: join B to C — triangle ABC is complete.
• Segment tool: join A to D.
• Segment tool: join B to D — triangle ABD is complete.
• Perp tool: click C, then click on AB — the height of triangle ABC.
• Perp tool: click D, then click on AB — the height of triangle ABD. It equals the height of ABC (same parallels), so the areas are equal.

The proof, step by step

Prove that triangles on the same base and between the same parallels are equal in area.

• Complete parallelograms ABEC and ABFD.
• BC and BD are diagonals, so each triangle is half its parallelogram.
• Parallelograms on same base and parallels are equal in area.
• Therefore, Area(ΔABC) = Area(ΔABD).

Worked example

In the figure, ΔABC and ΔABD share base AB. The distance between AB and the line CD is 8 cm. If AB = 12 cm, find the area of ΔABC.

Both triangles share the same base AB and lie between the same parallels, so their heights are equal (8 cm). Area = ½ × 12 × 8 = 48 cm².

• 48 cm² — correct
• 96 cm²
• 24 cm²
• 36 cm²