Given a right trapezoid ABCD with side lengths AB=2,BC=CD=10,AD=6, a circle is drawn through points B and D, intersecting the extension of BA at point E and the extension of CB at point F. Then the value of BE−BF is
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
As shown in Figure 3, extend CD to intersect ⊙O at point G. Let the midpoints of BE and DG be M and N, respectively. It is easy to see that AM=DN. Since BC=CD=10, by the secant theorem, it is easy to prove that BF=DG. Therefore, BE−BF=BE−DG=2(BM−DN)=2(BM−AM)=2AB=4.
Source: NuminaMath-1.5,
licensed Apache-2.0.
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