Question

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

Answer 1

Given A ∆ ABC in which /_ ABC > 90° and AD perpendicular ( CB produced ).


To prove :- AC² = AB² + BC² + 2 • BC • BD .


Proof :-


In ∆ ADB , /_ ADB = 90°

Therefore,


AB² = AD² + BD² ---------(1) [ By Pythagoras theroem]


In ∆ ADC , /_ ADC = 90°

Therefore,


AC² = AD² + CD² [ By Pythagoras theroem ]


=> AD² + ( BC + BD )² [ Since CD = ( BC + BD ) ]



=> AD² + ( BC² + BD² + 2•BC • BD )



=> ( AD² + BD² ) + ( BC² + 2 • BC • BD )


=> ( AB² + BC² + 2 BC BD ) [ Using (1) ]

Answer 2

ⓗⓔⓨ ⓕⓡⓘⓔⓝⓓ

ⓗⓔⓡⓔ ⓘⓢ ⓨⓞⓤⓡ

ⓐⓝⓢⓦⓔⓡ



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⊙ In triangle ADB .. ADB = 90°


⊙ so, AB^2 = AD^2 + BD^2 .... Pythagoras theorm


In triangle ADC...
⊙


AC^2 = AD^2 + CD^2 ..... by Pythagoras theorm.



⊙ AD^2 + ( BC + BD ) ^2 ( since , CD = BC+BD )



⊙ AD^2 + ( BC^2 + BD^2 + 2* BC*BD )


⊙ AD^2 + BD^2 +( BC^2 +2 *BC*BD )


⊙ AB^2 + BC^2 + 2 BC * BD


$hope \: it \: helps$





$<marquee > prabhudutt$