Question

In the adjoining figure, angle B of ∆ABC is an acute angle and AD is perpendicular to BC. prove that
AC² = AB²+BC²-2BC×BD

Answer 1

$\underline\bold\pink{\huge{HELLO\:DEAR}}$




$\underline\bold\red{\huge{ANSWER : }}$

In ∆ABD, angle ADB = 90°

So, AB²=AD²+BD²

Again, In ∆ADC, angle ADC = 90°.

So, AC²=AD²+DC²
=> AC²=AD²+(BC-BD) ²
=> AC²=AD²+BC²+BD²-2×BC×BD
=> AC²=(AD²+BD²) +BC²-2×BC×BD
=> AC²=AB²+BC²-2BC×BD [PROVED]