Q. 194.1( 9 Votes )

In the given figu

Answer :

Given: AD CB (produced)

To prove: AC2 = AB2 + BC2 + 2BC • BD


In ∆ ADC, DC = DB + BC ……….(i)


First, in ∆ ADB,


Using Pythagoras theorem, we have


AB2 = AD2 + DB2 AD2 = AB2 – DB2 ……….(ii)


Now, applying Pythagoras theorem in ∆ ADC, we have


AC2 = AD2 + DC2


= (AB2 – DB2) + DC2 [Using (ii)]


= AB2 – DB2 + (DB + BC)2 [Using (i)]


Now, (a + b)2 = a2 + b2 + 2ab


AC2 = AB2 – DB2 + DB2 + BC2 + 2DB • BC


AC2 = AB2 + BC2 + 2BC • BD


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