Question

the perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB = 3CD prove that 2AB^2 = 2AC^2 + BC^2

Answer 1

Answer:

Step-by-step explanation:

Given in ΔADB, AB²=AD²+BD².................eq.1

           In ΔADC, AC²=AD²+CD²...................eq.2

Subtracting eq.1 and eq.2,

                         AB²-AC²=AD²+BD²-AD²-CD²

                                        =BD²-CD²

                                        =3/4BC²-1/4BC²

                                        =2/4 BC²

                                        =1/2 BC²

                  2(AB²-AC²)=BC²

                =2AB²-2AC²=BC²

                =2AB²=BC²+2AC²

Hence,proved.

               

                               

Answer 2

Answer:

See the attachment. Hope it helps uh