Concyclic Points from Midpoint of an Arc

What is this about?


In circle $A(B)$, centered at $A$ and passing through $B$, $B$ is the midpoint of arc $CC'.$ $D$ and $E$ are points on the circle. $BD$ cuts $CC'$ in $F,$ $BE$ in $G.$

Midpoint of an arc and concyclic points

Prove points $D,$ $E,$ $F,$ $G$ concyclic.


Think of inscribed, secant angles and angle chasing.


To prove the required concyclicity suffice it to show that $\angle DFG+\angle DEG=180^{\circ}.$

Midpoint of an arc and concyclic points - solution

In terms of arcs,

$ \begin{align} \angle DFG &= \angle DFC' \\ &=({DEC'}+{BC})/2 \\ \end{align} $


$ \begin{align} \angle DEG &= \angle DEB \\ &=({CD}+{BC})/2 \\ &=({CD}+{C'B})/2 \\ \end{align} $

The arcs involved cover the whole circle and, hence, add up to $360^{\circ}.$


The problem is from the Geometry in Pictures by A. V. Akopyan. The book is in Russian but this should not scare anyone, for it is a collection of suggestive geometric diagrams with no description, let alone description in Russian.

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