Angle chasing to show three points are collinear.Let ABC be an acute triangle with circumcenter O and let K be such that KA is tangent to the circumcircle of △ A B C and ∠ K C B = 90 ∘ . Point D lies on BC such that KD||AB. Show that DO passes through A.

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Angle chasing to show three points are collinear.Let ABC be an acute triangle with circumcenter O and let K be such that KA is tangent to the circumcircle of   △  A  B  C and   ∠  K  C  B  =      90          ∘      . Point D lies on BC such that KD||AB. Show that DO passes through A.

Siena Bennett · Accepted Answer

Step 1We shall try to work backwards in order to reach the desired conclusion.Let   △  A  B  C be an acute triangle with circumcenter O and let K be such that KA is tangent to the circumcircle of   △  A  B  C and   ∠  K  C  B  =      90          ∘        ..Define L to be any point on line KA such that A lies between K and L. Also, extend segment AO to point X on BC.Step 2Using Alternate Segment Theorem,   ∠  B  A  L  =  A  C  B  =  α    ⟹    ∠  A  O  B  =  2  α    ⟹    ∠  O  A  B  =  ∠  B  A  X  =  90  −  α  .Since,   ∠  X  A  L  =  ∠  X  C  K  =      90          ∘        ,  A  K  C  X is cyclic and hence,  ∠  A  K  X  =  ∠  A  C  X  =  α    ⟹    A  B  ∥  K  X  .But there exists a unique point D on BC such that AB∥KD. Therefore,   X  ≡  D and A, O, D are collinear.

Lokubovumn · Accepted Answer

Step 1It is sufficient to show that   ∠  C  A  O  =  ∠  C  A  D.Let the tangent be labelled KAK′. Then due to the Alternate Segment Theorem,   ∠      K    ′    A  B  =  ∠  A  C  B.And since AB∥KC, both of these equal   ∠  A  K  D. The tangent is perpendicular to the radius so   ∠  B  A  O  =      90    o    −  ∠      K    ′    A  B.Step 2But from the data given,   ∠  K  C  A  =      90    o    −  ∠  A  C  B    ⟹    ∠  K  C  A  =  ∠  B  A  O.Considering the total angle in triangles ABC and AKC, the remaining angle in each is the same, so   ∠  O  A  C  =  ∠  D  K  C.However, points AKC and D are concyclic since   ∠  A  K  D  =  ∠  A  C  D (due to Angles in the Same Segment are Equal).Therefore, for the same reason,   ∠  D  A  C  =  ∠  D  K  C.Hence, finally,   ∠  C  A  O  =  ∠  C  A  D

Let ABC be an acute triangle with circumcenter O and let K be such that KA is tangent to the circumcircle of triangle ABC and angle KCB=90^circ. Point D lies on BC such that KD||AB. Show that DO passes through A.

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