# math vectors triangle A B C modulus AB +AC =mod AB -AC M Show triangle AB C right

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You need to use the information provided by the problem such that:

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

You need to expand the squares both sides, such that:

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

Reducing duplicate terms 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`

Since the dot product of the vectors `bar(AB)` and `bar(AC)` is 0, then the vectors are orthogonal to each other.

**Since `AB _|_ AC` , yields that triangle `Delta ABC` is a right angle triangle.**