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Bulgarian Mathematical Competitions

Geometry Difficulty 4.7 AIME Prove it Bulgaria

Problem:
Let $k$ be the circumcircle of $\triangle ABC$ with $\angle ACB > 90^\circ$ , and $BD$ be the diameter of $k$ through $B$ . The circle $k_1$ with center $D$ and radius $DC$ meets $k$ at point $E$ and $AB$ at point $G$ . If $F$ is the intersection point of $GE$ and $BD$ , prove that $\angle DCG = \angle EFD$ .

Solution

Solution:
Since $\angle ACB > 90^\circ$ , the points $G$ and $E$ lie on different sides to the diameter $BD$ . If $\angle DCG = \angle EFD$ , then $\angle DCG + \angle DFG = 180^\circ$ , i.e. it is enough to prove that the quadrilateral $CDFG$ is cyclic. On the other hand, $CE \perp BD$ and $D$ is the center of $k_1$ . Therefore $\angle CDF = \frac{1}{2} C\overset{\frown}{G}_{k_1}$ (as a central angle), $\angle CGF = \angle CGE$ is inscribed in $k_1$ and $\angle CGF = \frac{1}{2}(360^\circ - C\widehat{G}E_{k_1})$ .

Then $\angle CDF + \angle CGF = 180^\circ$ , i.e. the quadrilateral $CDFG$ is cyclic, whence $\angle DCG = \angle EFD$ .

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