I have been posed the following conjecture:
Let $AB$ be the diameter of a circle.
Let $C$ be the mid point of the arc $AB$.
Let $D$ be somewhere on the arc $AC$ and $E$ be on the chord $DB$ such that $AD=EB$.
Is it true that angle $ECD$ is always a right angle?
2 Answers
$\begingroup$Join $AC$ and $BC$. Clearly $\angle ACB$ is a right angle, so we need to show $\angle ACD=\angle BCE$.
In triangles $\triangle ACD$ and $\triangle BCE$ we have
- $AD=BE$ (given);
- $AC=BC$ ($C$ is the midpoint of arc $AB$);
- $\angle DAC=\angle EBC$ (angles at circumference standing on the chord $DC$);
so the triangles are congruent (SAS), $\angle ACD=\angle BCE$ and we are finished.
$\endgroup$ $\begingroup$As shown in the diagram, add the lines $AC$ and $BC$, and extend the line $CE$ to $F$.
Note that $AC = BC$ by hypothesis and $\angle DAC = \angle DBC$ (as both angles inscribe the same arc); this plus the hypothesis $AD = BE$ gets us the triangle congruence $\triangle ACD = \triangle BCE$, from which it follows that $\angle DCA = \angle BCF$. Equal angles inscribe equal arcs, so the arcs $AD$ and $BF$ are equal; it follows that $\angle DCF$ inscribes an arc of $180^\circ$ (the half-circle $AFB$, minus $BF$, plus $AD$), so it is a right angle.
$\endgroup$
