Question

In ΔABC, AD is perpendicular to BC. Prove that: AB² + CD² = AC² + BD²

Answer 1

Since triangle ABD and ACD are right triangle right-angled at D,
AB² = AD² + BD²    ....(i)
AC² = AD² + CD²   ....(ii)
Subtracting (ii) and (i), we get
AB² - AC² = BD² - CD²
⇒ AB² + CD² = AC² + BD².