Question

ABC is a triangle . angle b of triangle abc is an acute angle and ad perpendicular bc . prove that AC^2 = AB^2 +BC^2 - 2BC ×BD​

Answer 1

Answer:

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

In $\triangle$ ABD,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

AB$^{2}$ = AD$^{2}$ + BD$^{2}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

In $\triangle$ ADC

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

AC$^{2}$ = AD$^{2}$ + CD$^{2}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

AC$^{2}$ = AD$^{2}$ + ( BC – BD$^{2}$ )

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

⟹ AC$^{2}$ = AB$^{2}$ – BD$^{2}$ + BC$^{2}$ + BD$^{2}$ – 2BC × BD

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

⟹ AC$^{2}$ = AB$^{2}$ + BC$^{2}$ – 2BC × BD.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Hence, proved.