Question

In a ∆ABC , angle <ABC < 90° and AD perpendicular CB . prove that AC² = AB² + BC² -2BC BD.​

Answer 1

Answer:

AC^2=AB^2+BC^2-2BC.BD

step by step solution

Answer 2

Given :-

• In ΔABC, angleABC > 90°

• AD perpendicular to CB

To prove :-

AC² = AB² + BC² - 2BC.BD

Proof :-

In ΔADB , AngleADB = 90°

By using Pythagoras theorem,

AB² = AD² + BD² ...eq( 1 )

Now,

In ΔADC, AngleADC = 90°

AC² = AD² + CD²

AC² = AD² + ( BD + BC)² [ CD = BD + BC ]

[ By using identity

( a + b)² = a² + b² + 2ab ]

AC² = AD² + BD² + BC² - 2BC.BD

AC² = (AD² + BD)²+ BC² - 2BC.BD

From eq( 1 )

AC² = AB² + BC² - 2BC.BD $.$

Hence, Proved