252. Same Height, Different Base · Area ratio follows the base ratio

Try to break it

Drag B along base AC. △ABD and △BCD share the same apex D, so they share the same height. As B moves, the base lengths AB and BC change in opposite directions, and the ratio of their areas always equals AB / BC. Try to find a position where the area ratio drifts away from the base ratio; impossible.

How you build it

Construct two triangles on the same base with same vertex.

• Point tool: mark A — the left end of the base.
• Point tool: mark C — the right end of the base.
• Segment tool: join A to C — the common base line.
• Point tool: click on AC to place B between A and C — it splits the base into AB and BC.
• Point tool: mark D above the base — the apex shared by both triangles.
• Segment tool: join D to A.
• Segment tool: join D to B — this side splits the figure into triangles ABD and BCD.
• Segment tool: join D to C — triangles ABD and BCD share the apex D.
• Perp tool: click D, then click on AC — drop the perpendicular. Its foot on AC is H, and DH is the height shared by both triangles.
• Point tool: mark H where the perpendicular meets AC — the common foot of the height DH.

The proof, step by step

Prove that triangles with the same height have areas in the ratio of their bases.

• Drop perpendicular DH from vertex D to the base line AC. This segment DH is the common height for both ΔABD and ΔBCD.
• Write the area formulas: Area(ΔABD) = ½ × AB × DH and Area(ΔBCD) = ½ × BC × DH.
• Divide the two area expressions. The factors ½ and DH cancel out, proving Area(ΔABD)/Area(ΔBCD) = AB/BC.

Worked example

In the figure, AB = 6 cm and BC = 4 cm. If the area of ΔABD is 18 cm², find the area of ΔBCD.

Since they share the same height, Area(ABD)/Area(BCD) = AB/BC = 6/4 = 3/2. So 18/Area(BCD) = 3/2 ⇒ Area(BCD) = 12 cm².

• 10 cm²
• 12 cm² — correct
• 15 cm²
• 24 cm²