If K is the midpoint of AH, P ∈ A B, Q ∈ A C and K ∈ P Q such that O K ⊥ P Q , then O P = O Q

Question

If K is the midpoint of AH,   P  ∈  A  B,   Q  ∈  A  C and   K  ∈  P  Q such that   O  K  ⊥  P  Q , then   O  P  =  O  Q

heilaritikermx · Accepted Answer

Step 1Let D be the intersection of AO with the circumcircle. So, we have BHCD is a parallelogram, and thus BC bisects HD. Let M be the intersection of BC and HD, so M is the midpoint of HD. Let EF be the perpendicular line of HD pass H. Since   E  F  ⊥  H  D ,   B  H  ⊥  A  C , so   ∠  B  H  C  =  ∠  A  F  E . Also, since   B  D  ⊥  A  B ,   E  F  ⊥  D  H , so   ∠  B  D  H  =  ∠  A  E  F . Thus,   △  A  E  F  ∼  △  B  D  H . Also, since M on HD and   ∠  H  B  M  =      90    ∘    −  ∠  A  C  B  −  ∠  F  A  H , we also have   △  B  H  M  ∼  △  A  F  H . So   H  F      /    F  E  =  H  F      /    A  F  ×  A  F      /    F  E  =  M  H      /    B  H  ×  B  H      /    H  D  =  H  M      /    H  D  =  1      /    2 . Therefore,   E  F  =  2  F  H , so   E  H  =  F  H . Since K is the midpoint of AH, O is the midpoint of AD, we have   H  D  ∥  O  K . Also   O  K  ⊥  P  Q and   H  D  ⊥  E  F , we have   P  Q  ∥  E  F . So   P  K      /    K  Q  =  E  H      /    H  F  =  1 . Thus,   P  K  =  K  Q . Also, since   O  K  ⊥  P  Q , we can arrive at   O  P  =  P  Q .

If K is the midpoint of AH, P ∈ A B , Q ∈ A C and K

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