Try this beautiful Problem on Geometry based on Measure of angle from AMC 10 A, 2014. You may use sequential hints to solve the problem.
Let $\triangle A B C$ be an isosceles triangle with $B C=A C$ and $\angle A C B=40^{\circ} .$ Construct the circle with diameter $\overline{B C}$, and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{A C}$ and $\overline{A B}$, respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $B C D E .$ What is the degree measure of $\angle B F C ?$
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Geometry
Circle
Triangle
Pre College Mathematics
AMC-10A, 2019 Problem-`13
$110^{\circ}$
According to the questation we draw the diagram. we have to find out \(\angle BFC\)
Now \(\angle BEC\) = \(\angle BDC\) =\(90^{\circ}\) (as they are inscribed in a semicircle)
$\angle A C B=40^{\circ} .$ Therefore we can say that \(\angle ABC=70^{\circ}\) (as $\triangle A B C$ be an isosceles triangle with $B C=A C$)
Can you find out the value of $\angle B F C ?$
Now can you finish the problem?
As \(\angle ABC=70^{\circ}\) and \(\angle BEC=90^{\circ}\) Therefore $\angle E C B=20^{\circ}$( as sum of the angles of a triangle is\( 180^{\circ}\)
Similarly $\angle D B C=50^{\circ}$
Now Can you finish the Problem?
Now $\angle B D C+\angle D C B+\angle D B C=180^{\circ} \Longrightarrow 90^{\circ}+40^{\circ}+\angle D B C=180^{\circ} \Longrightarrow \angle D B C$=$50^{\circ}$
$\angle B E C+\angle E B C+\angle E C B=180^{\circ} \Longrightarrow 90^{\circ}+70^{\circ}+\angle E C B=180^{\circ} \Rightarrow \angle E C B$=$20^{\circ}$
we take triangle $B F C$, and find $\angle B F C=180^{\circ}-50^{\circ}-20^{\circ}=110^{\circ}$
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