In a Triangle ABC, and D is the mid point of AC.The value of BC^2-BD^2 = AD^2.Prove.

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- Aug 20th 2008, 12:25 AMdeviTriangles
In a Triangle ABC, and D is the mid point of AC.The value of BC^2-BD^2 = AD^2.Prove.

- Aug 20th 2008, 01:34 AMticbol
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. - Aug 20th 2008, 08:40 AMSoroban
Hello, devi!

The problem has strange wording ... and the statement is**not true**.

Quote:

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: .

- Aug 20th 2008, 09:12 AMIan1779
Triangle BCD is not right-angled so you cannot apply Pythagoras Theorem to it.