# Vectors AB + AC and AB - AC have same modulus then prove triangle ABC is right ?

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The problem provides the information that the absolute values of the vectors `bar(AB) + bar(AC)` and `bar(AB) - bar(AC)` coincide, hence, the squares of the absolute values of the vectors `bar(AB) + bar(AC)` and `bar(AB) - bar(AC)` also coincide, such that:

`|bar(AB) + bar(AC)| = |bar(AB) - bar(AC)| => |bar(AB) + bar(AC)|^2 = |bar(AB) - bar(AC)|^2 => (bar(AB))^2 + 2bar(AB)*bar(AC) + (bar(AC))^2 = (bar(AB))^2 - 2bar(AB)*bar(AC) + (bar(AC))^2`

Reducing duplicate members both sides, yields:

`2bar(AB)*bar(AC) = -2bar(AB)*bar(AC)`

`2bar(AB)*bar(AC) + 2bar(AB)*bar(AC) = 0`

`4bar(AB)*bar(AC) = 0 => bar(AB)*bar(AC) = 0`

You need to remember that if the dot product of two vectors yields 0, then the vectors are perpendicular, such that:

`bar(AB)*bar(AC) = 0 => bar(AB)_|_bar(AC) => hat A = 90^o`

**Hence, since the vectors `bar(AB),bar(AC) ` are perpendicular, yields that the angle `hat A = 90^o` , thus the triangle `ABC` is a right triangle.**