A NECESSARY_ and_SUFFICIENT condition of tangency to the circle

               Given a triangle $ABC$,

          a necessary and sufficient condition for the line $CD$ to be tangent

          to the circle circumscribing the triangle is $\angle{BAC} \equiv \angle{BCD}.$

 

          The condition is necessary : $CD$ - is tangent to circle $\Rightarrow\;\angle{BAC}=\angle{BCD}.$

     Indeed, we have the equalities between the measures of angles and arcs of the circle:

$$\widehat{BAC}=\frac{\overset{\frown}{BC}}{2},\;\;\widehat{BCD}=\frac{\overset{\frown}{BC}}{2}$$
hence the conclusion.

          The condition is sufficient : $\angle{BAC}=\angle{BCD}\;\Rightarrow$ $CD$ - is tangent to the

                                            circle circumscribing the triangle.

          For demonstration we take $O$ the center of the circumscribed circle. 

          If the point $O$ is inside the triangle, then we write the relation

$$\widehat{AOB}+\widehat{BOC}+\widehat{COA}=180^{\circ}.$$

          We show the proof when point $O$ is outside the triangle. In this case we have the relationship

$$\widehat{BOA}+\widehat{AOC}=\widehat{BOC}.$$

     From the isosceles triangles $AOB\;(OA=OB)$ and $AOC\;(OA=OC)$ it follows

$$\widehat{BOA}=180^{\circ}-2 \cdot \widehat{BAO},\;\;\widehat{AOC}=180^{\circ}-2 \cdot \widehat{CAO}$$

 so         $\widehat{BOC}=360^{\circ}-2 \cdot (\widehat{BAO}+\widehat{CAO})=360^{\circ}-2 \cdot \widehat{BAC}.$

But in the isosceles triangle $BOC$ we have

$$\widehat{BCO}=\frac{180^{\circ}-\widehat{BOC}}{2}=\frac{2\cdot \widehat{BAC}-180^{\circ}}{2}=\widehat{BAC}-90^{\circ}.$$

 Using the hypothesis $\widehat{BAC}=\widehat{BCD}$ results $$\widehat{OCD}=\widehat{BCD}-\widehat{BCO}=\widehat{BAC}-(\widehat{BAC}-90^{\circ})=90^{\circ},$$

so the line $CD$ is perpendicular to the radius $OC$ of the circle circumscribing the triangle $ABC$, i.e. it is tangent to this circle.

$\blacksquare$

               As an application of this property we solve the following problem:

(from the GMB exercise supplement, from October 2009 (page 3; proposed for 7th grade at a student summer camp). In translation, thanks to Mr DeepL

          "The median $AM$ of the acute triangle $ABC$ intersects the circle

circumscribing the triangle a second time at point $D$. If $E$ is the symmetric 

point of $A$ with respect to $M$, then show that the line $BC$ is the common 

tangent of the circumscribed circles of triangles $BDE$ and $CDE$."

          For demonstration, let's take a simplified figure.

From $MB=MC$ and $MA=ME\;\;\Rightarrow\;BACE$ - parallelogram. We have equals angles:

$$\angle{CAE} \equiv \angle{AEB}\;\;\;\angle{BAE}\equiv \angle {AEC},$$

 or written 

$$\widehat{CAD}=\widehat{DEB},\;\;\;\widehat{BAD}=\widehat{DEC}.\tag{1}$$

 We still have the equalities between angles and arcs:

$$\widehat{CAD}=\frac{\overset{\frown}{CD}}{2}=\widehat{CBD,}\;\;\;\widehat{BAD}=\frac{\overset{\frown}{BD}}{2}=\widehat{BCD}. \tag{2}$$

     From (1) and (2) follows

$\widehat{DBC}=\widehat{DEB}$    hence $BC$ is tangent to the circle circumscribed of $\Delta BDE,$

$\widehat{DCB}=\widehat{DEC}$     hence $CB$ is tangent to the circle circumscribed of $\Delta CDE$.

$\blacksquare\;\blacksquare$