2019 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 13 Medium

Let \triangle ABC be an isosceles triangle with BC=AC and \angle ACB=40^\circ. Construct the circle with diameter \overline{BC}, and let D and E be the other intersection points of the circle with the sides \overline{AC} and \overline{AB} , respectively. Let F be the intersection of the diagonals of the quadrilateral BCDE. What is the degree measure of \angle BFC? (2019 AMC 10A Problem, Question#13)

  • A.

    90

  • B.

    100

  • C.

    105

  • D.

    110

  • E.

    120

Answer:D

Drawing it out, we see \angle BDC and \angle BEC are right angles, as they are inscribed in a semicircle.Using the fact that it is an isosceles triangle, we find \angle ABC=70^\circ. We can find \angle ECB=20^\circ and \angle DBC=50^\circ by the triangle angle sum on \triangle ECB and \triangle DBC .

\angle BDC+\angle DCB+\angle DBC=180^\circ\Rightarrow90^\circ+40^\circ+\angle DBC=180^\circ\Rightarrow\angle DBC=50^\circ

\angle BEC+\angle EBC+\angle ECB=180^\circ\Rightarrow90^\circ+70^\circ+\angle ECB=180^\circ\Rightarrow\angle ECB=20^\circ

Then, we take triangle BFC, and find \angle BFC=180^\circ-50^\circ-20^\circ=110^\circ

Alternatively, we could have used similar triangles. We start similarly to Solution 1.Drawing it out, we see \angle BDC and \angle BEC are right angles, as they are inscribed in a semicircle.Therefore, \angle BDA=180^\circ-\angle BDC=180^\circ-90^\circ=90^\circ.

So, \triangle BEF\sim BDA by AA Similarity, since \angle EBF=\angle DBA and \angle BEC=90^\circ=\angle BDA.Thus, we know \angle EFB=\angle DAB=\angle CAB=70^\circ, Finally, we deduce\angle BFC=180^\circ-\angle EFB=180^\circ-70^\circ=110^\circ.

Through the property of angles formed by intersecting chords, we find that m\angle BFC=\dfrac{m\overset{\frown}{BC}+m\overset{\frown}{DE}}{2}

Through the Outside Angles Theorem, we find that m\angle CAB=\dfrac{m\overset{\frown}{BC}-m\overset{\frown}{DE}}{2}

Adding the two equations gives us m\angle BFC-m\angle CAB=m\overset{\frown}{BC}\Rightarrow m\angle BFC=m\overset{\frown}{BC}-m\angle CAB

Since \overset{\frown}{BC} is the diameter, m\overset{\frown}{BC}=180^\circ, and because \triangle ABC is isosceles and m\angle ACB=40^\circ, we have m\angle CAB=70^\circ. Thus m\angle BFC=180^\circ-70^\circ=110^\circ.

Notice that if \angle BEC=90^\circ, then \angle BEC and \angle ACE must be 20^\circ. Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that \angle EBDā‰Œ\angle ECD=20^{\circ}.Thus \angle CBF=70-20=50^\circ, and so \angle BFC=180-20-50=110^\circ, which is \text{D}.

Note: As in many elementary geometry problems, if you can't see how to solve it, you could simply draw an accurate diagram and measure the angle using a protractor as 110^\circ.

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