Question

In triangle ABC (Figure 3), AD perpendicular BC. Prove that
AC^2 = AB^2 + BC^2 - 2BC X BD​

Answer 1

Answer:

AC^2 = AB^2 + BC^2 - 2BC.BD

Step-by-step explanation:

In ∆ABD : Using Pythagoras theorem :

= > AB^2 = AD^2 + BD^2 ...( 1 )

In ∆ADC : Using Pythagoras theorem :

= > AC^2 = AD^2 + DC^2

= > AC^2 = AD^2 + ( BC - BD )^2 { BC - BD = DC }

= > AC^2 = AD^2 + BC^2 + BD^2 - 2BC.BD { ( a - b )^2 = a^2 + b^2 - 2ab }

= > AC^2 = AD^2 + BD^2 + BC^2 - 2BC.BD

= > AC^2 = AB^2 + BC^2 - 2BC.BD { from ( 1 ) }

Hence proved.