Question

please please give answer quickly. tomorrow having exam . pls answer the question fast. please

Answer 1

Solution :

Given :

In ∆ABC , AD perpendicular to BC,

<B < 90°

To prove : AC² = AB² + BC² - 2BC×BD

proof :

i ) In ∆ADB , <ADB = 90°

By Phythogarian theorem ,

AB² = AD² + BD² ----( 1 )

ii ) In ∆ADC , <ADC = 90°

AC² = AD² + CD²

= AD² + ( BC - BD )²

[ Since , CD = BC - BD ]

= AD² + BC² + BD² - 2×BC×BD

= ( AD² + BD² ) + BC² - 2×BC×BD

= AB² + BC² - 2BC×BD [ from ( 1 ) ]

Hence proved.

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