Question

in triangle ABC AD perpendicular to BC and BD:CD=3:1 prove that 2(AB^2-AC^2)=BC^2

Answer 1

hey dear!
here is ur answer
.........
let BD=3x n CD=x
in TriABD . by phy th
AB^2=BD^2+AD^2
AD^2=AB^2-BD^2......let this eq 1
now in triACD .BY Phy th
AC^2=AD^2+CD^2
AD^2=AC^2-CD^2....let this eq 2
from 1 n 2
we get
AB^2-BD^2=AC^2-CD^2
AB^2-AC^2=BD^2-CD^2
AB^2-AC^2=9x^2 - x^2
AB^2-AC^2=8X^2
AB^2-AC^2=2 (4x^2)
AB^2-AC^2=2 (BD/2)^2
AB^2-AC^2=2BC^2/4
AB^2-AC^2=BC^2/2
2 (AB^2-AC^2)=BC^2

hope this will help you :)

Answer 2

Answer :

Step-by-step explanation:

Given: ∆ABC, AD perpendicular BC,BD:CD =3:1

To prove : 2(AB^2-AC^2) =BC^2

Proof: let BD be 3x and CD be x

in ∆ABD by phythagorous ∆le

AB^2=BD^2+AD^2

AD^2 =AB^2-BD^2 eqn 1

now in ∆ACD by phythagorous

AC^2= AD^2+CD^2

AD^2 = AC^2 - CD^2 eqn 2

from 1 & 2

we get AB^2-BD^2 = AC^2-CD^2

AB^2-AC^2= BD^2- CD^2

AB^2-AC^2 =9x^2-x^2

AB^2-AC^2 =8^2

AB^2 - AC^2=2(4x^2)

AB^2-AC^2= 2(BD/2)^2

AB^2- AC^2=2BD^2/4

AB^2-AC^2 = 2BC^2/2

2(AB^2-AC^2) = BC^2

Hence proved