In circle K shown, secant PEC intersects chord BKD at F. If AE = AB, mED = 32, and mBC mCD = 2 :1, then find the following:
1 answer:
Answer:
(a) ∠P = 60°
(b) ∠CFD = 104°
(c) no
Step-by-step explanation:
It is convenient to start by finding the measures of the arcs around the circle. We know BD is a diameter, so arc BD measures 180°. Point C divides it into parts with the ratio 2:1, so ...
arc BC = (2/3)(180°) = 120°
arc CD = (1/3)(180°) = 60°
Arc BE is the supplement to arc ED, so is ...
arc BE = 180° -32° = 148°
and arcs BA and AE are each half that, so are ...
arc BA = arc AE = 148°/2 = 74°
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(a) Arc ABC = arc BA +arc BC = 74° +120° = 194°
Arc AE = 74° (from above). The angle at P is half the difference of arcs ABC and AE:
∠P = (194° -74°)/2
∠P = 60°
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(b) ∠CFD is half the sum of arcs CD and BE:
∠CFD = (60° +148°)/2
∠CFD = 104°
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(c) ∠CEA is half the measure of arc CBA, so is ...
∠CEA = 194°/2
∠CEA = 97°
AE is not perpendicular to PC.
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The attached diagram is to scale.
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