Problem:
Given regular pentagon ABCDE, a circle can be drawn that is tangent to DC at D and to AB at A. The number of degrees in minor arc AD is
Answer Choices:
A. 72
B. 108
C. 120
D. 135
E. 144
Solution:
Let O be the center of the circle. Since the sum of the interior angles in any n-gon is ( n−2 ) 180∘, the sum of the angles in ABCDO is 540∘. Since ∠ABC=∠BCD=108∘ and ∠OAB=∠ODC=90∘, it follows that the measure of ∠AOD, and thus the measure of minor arc AD, equals 144∘.
OR
Draw AD. Since △AED is isosceles with ∠AED=108∘, it follows that ∠EDA=∠EAD=36∘. Consequently, ∠ADC=108∘−36∘=72∘. Since ∠ADC is a tangentchord angle for the arc in question, the measure of the arc is 2(72∘)=144∘.
OR
Let O be the center of the circle, and extend DC and AB to meet at F. Since ∠DCB=108∘ and △BCF is isosceles, it follows that ∠AFD=[180∘−2(180∘−108∘)]=36∘. Since ∠ODF=∠OAF=90∘, in quadrilateral OAFD we have angles AOD and AFD supplementary, so the measures of angle AOD and the minor arc AD are 180∘−36∘=144∘.
Note. A circle can be drawn tangent to two intersecting lines at given points on those lines if and only if those points are equidistant from the point of intersection of the lines.
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