Let ABC be a triangle such that ∠ A C B is acute. Suppose that D is an interior point of the triangle ABC such that ∡ A D B = ∡ A C B + π 2 and A C ⋅ B D = A D ⋅ B C ..a) Find A B ⋅ C D A C ⋅ B D ..(b) Show that the tangents at C to the circumcicle of the triangle ACD and the circumcircle of the triangle BCD are perpendicular.

Question

Let ABC be a triangle such that   ∠  A  C  B is acute. Suppose that D is an interior point of the triangle ABC such that   ∡      A    D    B    =  ∡      A    C    B    +      π    2   and   A  C  ⋅  B  D  =  A  D  ⋅  B  C    ..a) Find             A      B      ⋅      C      D              A      C      ⋅      B      D          ..(b) Show that the tangents at C to the circumcicle of the triangle ACD and the circumcircle of the triangle BCD are perpendicular.

Urijah Hahn · Accepted Answer

Step 1Draw the perpendicular to CB and then choose E on it, s.t.   C  B  =  C  E, as in the picture below. Now obviously   ∠  A  C  E  =  ∠  A  D  B and also from the condition:            A      C              C      E        =            A      C              B      C        =            A      D              D      B          .Step 2Therefore   △  A  C  E  ∼  △  A  D  B. So in particular we have   ∠  C  A  E  =  ∠  D  A  B. Also from the similarity of the triangles we have that                     A        C                    A        D              =                    A        E                    A        B                ..This gives us that   △  A  C  D  ∼  △  A  B  E. So using that BCE is a right isosceles triangle we have from   △  A  C  D  ∼  △  A  B  E that   C  D  ⋅  A  B  =  E  B  ⋅  A  D  =      2    B  C  ⋅  A  D  =      2    A  C  ⋅  B  D.Hence the ratio is       2  .

Ethen Blackwell · Accepted Answer

Step 1Let&#039;s invert about D eith arbitrary radius r&amp;gt;0. For any point X in the plane let X∗ be the image of X under the inversion. Then, we will rewrite all conditions in terms of A∗, B∗, C∗ and D.Firstly,   ∠  A  C  B  =  ∠  A  C  D  +  ∠  B  C  D  =  ∠  D      A          ∗            C          ∗        +  ∠  D      B          ∗            C          ∗       and   ∠  A  D  B  =  ∠      A          ∗        D      B          ∗      , so we have   ∠  A  D  B  =  ∠      A          ∗        D      B          ∗      .Hence,   ∠      A          ∗            C          ∗            B          ∗        =      π    2  .Step 2Secondly, recall that for any points M and N (other than D) we have       M          ∗            N          ∗        =            R      2              D      M      ⋅      D      N        ⋅  M  N     and     D      M          ∗        =            R      2              D      M        , and       M          ∗            N          ∗        =            R      2              D      M      ⋅      D      N        ⋅  M  N     and     D      M          ∗        =            R      2              D      M        ,, so the second equality can be rewritten as             R      2              D              A                  ∗                    ⋅      D              C                  ∗                      ⋅      A          ∗            C          ∗        ⋅            R      2              D              B                  ∗                      =            R      2              D              B                  ∗                    ⋅      D              C                  ∗                      ⋅      B          ∗            C          ∗                  R      2              D              A                  ∗                      ,, or       A          ∗            C          ∗        =      B          ∗            C          ∗        ..Thus, the triangle       A          ∗            B          ∗            C          ∗       is isosceles and right-angled with   ∠      A          ∗            C          ∗            B          ∗        =      π    2

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