# Triangles

• Aug 19th 2008, 11:25 PM
devi
Triangles
In a Triangle ABC, $|_A = 90 degree$ and D is the mid point of AC.The value of BC^2-BD^2 = AD^2.Prove.
• Aug 20th 2008, 12:34 AM
ticbol
Quote:

Originally Posted by devi
In a Triangle ABC, $|_A = 90 degree$ 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.
• Aug 20th 2008, 07:40 AM
Soroban
Hello, devi!

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

Quote:

In $\Delta ABC,\;\angle A = 90^o$ and D is the midpoint of AC.
The value of: . $BC^2-BD^2 \:= \:AD^2$ .
. . . "the value of..." ?
Prove.

Consider a 3-4-5 right triangle.
Code:

      B*         | *  *         |  *    *    5       3|    *        *         |      *          *         |        *              *         * - - - - - - * - - - - - - - *         A      2      D      2      C

We have: . $BC = 5,\;BD = \sqrt{13},\;AD = 2$

. . But: . $(5)^2 -(\sqrt{13})^2 \;\;{\bf{\color{red}\neq}}\;\;(2)^2$

• Aug 20th 2008, 08:12 AM
Ian1779
Triangle BCD is not right-angled so you cannot apply Pythagoras Theorem to it.