In a Triangle ABC, and D is the mid point of AC.The value of BC^2-BD^2 = AD^2.Prove.
If D is the midpoint of AC, then AD = DC....and AC = 2(AD)
In right triangle ABC,
(BC)^2 = (AB)^2 +(2(AD))^2
(BC)^2 = (AB)^2 +4(AD)^2 -----------(i)
In right triangle ABD,
(BD)^2 = (AB)^2 +(AD)^2 ----------(ii)
Eq,(i) minus Eq.(ii),
(BC)^2 -(BD)^2 = 3(AD)^2
Oops...it is not proven.
Hello, devi!
The problem has strange wording ... and the statement is not true.
Consider a 3-4-5 right triangle.In and D is the midpoint of AC.
The value of: . . . . . "the value of..." ?
Prove.Code:B* | * * | * * 5 3| * * | * * | * * * - - - - - - * - - - - - - - * A 2 D 2 C
We have: .
. . But: .