Question

In a rite triangle ABC, rite angled at B. D is any point on BC , then prove that AC2 = AD2 + DC2 +2BD.DC

Answer 1

triangle ABD is also right angle triangle 
therefore AB² + BD² = AD²
⇒AB² = AD² - BD²    .............(i)
now, AC² = AB² + BC²
⇒AC² = AD² - BD² + (BD + DC)²
⇒AC² = AD² - BD² + BD² + 2BC.BD + DC²
⇒AC² = AD² + DC² + 2BD.DC
hence proved

Answer 2

Answer:

In ∆ABD,

By Pythagoras theorem

-> AB² + BD² = AD²

AB² = AD² - BD² .....(1)

In ∆ABC,

By Pythagoras theorem

-> AC² = AB² + BC²

AC² = (AD² - DB²) + (BD + DC)² [ From (1) and BC = BD + DC ]

AC² = AD² - BD² + BD² + DC² + 2(DB)(DC)

AC² = AD² + CD² + 2(BD)(DC)

.......................Hence proved....................