Question

The perpendicular AD on the base BC in D such that BD=3CD . Prove that 2 AB2 = 2AC2 + BC2

Answer 1

Step-by-step explanation:

BD = 3CD

=> CD = 1/4BC

and BD=> 3/4BC

In ∆ABD, AB^2 + BD^2 + AD^2 .......(2)

In ∆ACD, AC^2 + CD^2 + AD^2 .......(3)

Subtracting (3) from (2), we get,

AB^2 - AC^2 = DB^2 - CD^2

= (3/4BC)^2 - (1/4BC)^2

= 9/16BC^2 - 1/16BC^2

= 1/2BC^2

=> 2AB^2 - 2AC^2 = BC^2

=> 2AB^2 = 2AC^2 +BC^2

hence proved ....

Answer 2

Answer:

Step-by-step explanation:22.TheperpendicularfromAonsideBCof∆ABCmeetsBCatDsuchthatDB=3CD.Provethat2AB2=

2AC2+BC2