Let △ A B C be a triangle of sides AB, AC and BC which has the following equations A B ≡ a 1 x + b 1 y + c 1 = 0 A C ≡ a 2 x + b 2 y + c 2 = 0 B C ≡ p x + q y + r = 0Then an angle ∠ A is acute if | a 1 b 1 p q | | p q a 2 b 2 | ( a 1 a 2 + b 1 b 2 ) &lt; 0And the angle ∠ A is obtuse if | a 1 b 1 p q | | p q a 2 b 2 | ( a 1 a 2 + b 1 b 2 ) &gt; 0But I absolutely have no idea how to prove this theorem.

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Let   △  A  B  C be a triangle of sides AB, AC and BC which has the following equations  A  B  ≡      a    1    x  +      b    1    y  +      c    1    =  0    A  C  ≡      a    2    x  +      b    2    y  +      c    2    =  0    B  C  ≡  p  x  +  q  y  +  r  =  0Then an angle   ∠  A is acute if      |                                        a            1                                                b            1                                                p                          q                      |        |                            p                          q                                                  a            2                                                b            2                                |    (      a    1        a    2    +      b    1        b    2    )  &amp;lt;  0And the angle   ∠  A is obtuse if      |                                        a            1                                                b            1                                                p                          q                      |        |                            p                          q                                                  a            2                                                b            2                                |    (      a    1        a    2    +      b    1        b    2    )  &amp;gt;  0But I absolutely have no idea how to prove this theorem.

engaliar0l · Accepted Answer

Step 1For this proof I will assume, naturally, that none of the lines are parallel and they do not all intersect in one point. I will also assume for the moment that   r  ≠  0 so this line does not pass through the origin.Consider two lines parallel to AB and AC, call them OP and OQ, whose equations, respectively are      a    1    x  +      b    1    y  =  0and      a    2    x  +      b    2    y  =  0These lines meet BC at points P and Q. Since triangles ABC and OPQ are similar with   ∠  A the same as   ∠  O, it is sufficient to derive the result for   ∠  O  =  θ instead.The angle   θ will be acute if   cos  ⁡  θ  &amp;gt;  0 and obtuse if   cos  ⁡  θ  &amp;lt;  0.Since  cos  ⁡  θ  =                              O          P                →            ⋅                        O          Q                →                            |            O      P              |                    |            O      Q              |              , ,so for the acute angle, we just require            O      P        →    ⋅            O      Q        →    &amp;gt;  0Solving simultaneously, we get the coordinates of P to be      (                  r                  b          1                                      a          1                q        −                  b          1                p              ,                  −        r                  a          1                                      a          1                q        −                  b          1                p              )  Correspondingly, the coordinates of Q are the same but will the suffix changed from 1 to 2.Therefore the required dot product is            r              b        1                            a        1            q      −              b        1            p        ⋅            r              b        2                            a        2            q      −              b        2            p        +            −      r              a        1                            a        1            q      −              b        1            p        ⋅            −      r              a        2                            a        2            q      −              b        2            p        =                    r        2            (              a        1                    a        2            +              b        1                    b        2            )                      |                                                            a                1                                                                    b                1                                                                        p                                      q                                      |                    |                                                            a                2                                                                    b                2                                                                        p                                      q                                      |            Noting that   r  ≠  0 , for the angle to be acute, therefore, we require      |                                        a            1                                                b            1                                                p                          q                      |        |                                        a            2                                                b            2                                                p                          q                      |    (      a    1        a    2    +      b    1        b    2    )  &amp;gt;  0  , ,with the inequality reversed for the obtuse angle.Note that this is the same formula as that provided by the OP, but expressed, in my opinion, in a better way.In the case that   r  =  0 , then we would translate all three lines so that the intersection point of the first two lines became the origin, and the same argument would apply

Let <mi mathvariant="normal">△ A B C be a triangle of sides AB, AC a

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