Two circles cut orthogonally in A and B. A diameter of one of the circles is drawn cutting the other circle in C and D. Prove that B C ⋅ A D = A C ⋅ B D.

Question

Two circles cut orthogonally in A and B. A diameter of one of the circles is drawn cutting the other circle in C and D. Prove that   B  C  ⋅  A  D  =  A  C  ⋅  B  D.

Leroy Cunningham · Accepted Answer

Step 1  ∠  O  A  C  =  ∠  A  D  C by Tangent-secant theorem, hence   △  O  A  D  ∼  △  O  C  A  ⇒            A      C              A      D        =            O      C              O      D        ..Step 2Similarly             B      C              B      D        =            O      C              O      D            , hence               B      C              B      D        =            A      C              A      D        ⇒  B  C  ⋅  A  D  =  A  C  ⋅  B  D      , QED.

grippeb9 · Accepted Answer

Step 1Orthogonality means that   O  B  ⊥  B      O    ′   so OB is tangent to circle with center at O′.Because of tangent chord property we see that (blue)   Δ  B  C  O  ∼  Δ  D  B  O    ⟹                      O        B                    O        D              =                    B        C                    B        D            Similary (red) we have   Δ  A  C  O  ∼  Δ  D  A  O    ⟹                      O        A                    O        D              =                    A        C                    A        D            .Step 2Since   O  A  =  O  B we are done.

Two circles cut orthogonally in A and B . A diameter of one of the circles is drawn cutting the other circle in C and D. Prove that BC⋅AD=AC⋅BD.

Open question

Answer & Explanation

New Questions in High school geometry