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swapan86
@swapan86
June 2021
1
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please please give answer quickly. tomorrow having exam . pls answer the question fast. please
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mysticd
Verified answer
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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plz answer the question...
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Answers & Comments
Verified answer
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.
••••