Question

ABC is a right triangle right angled at B. Let D & E be any points on AB and BC respectively. Prove that AE×AE+CD×CD=AC×AC+DE×DE

Answer 1

ΔABC is a right triangle, so by the Pythagorean theorem, (1) AB²+BC² = AC² ΔABE is a right triangle, so by the Pythagorean theorem, (2) AB²+BE² = AE². ΔDBC is a right triangle, so by the Pythagorean theorem, (3) DB²+BC² = CD². ΔDBE is a right triangle, so by the Pythagorean theorem, (4) DB²+BE² = DE². Add BE²+DB² to both sides of equation (1) AB²+BC²+BE²+DB² = AC²+BE²+DB² Rearrange the terms: AB²+BE²+DB²+BC² = AC²+DB²+BE² To make things easier to see, let's put parentheses around the first two terms on the left, the last two terms on the left, and the last two terms on the right: (5) (AB²+BE²)+(DB²+BC²) = AC²+(DB²+BE²) Using (2) we replace (AB²+BE²) in (5) by AE² Using (3) we replace (DB²+BC²) in (5) by CD² Using (4) we replace (DB²+BE²) in (5) by DE² And we end up with what we were to prove: AE²+CD² = AC²+DE²