angleeuclidean-geometrygeometryquadrilateraltriangles
For reference: The perpendicular bisectors of sides $AD$ and $CD$ of an $ABCD$ parallelogram intersect at
a point $M$ that belongs to $BC$. Find $\angle MAD$ if $\angle B = 110^\circ $ .
My progress
I drew some auxiliary lines but it wasn't enough
$\angle C = 180^\circ – \angle 110^\circ = \angle 70^\circ$
$\angle CMG = 180^\circ-160^\circ = 20^\circ$
The assumption is not necessary. Construct $P$ on the parallel to $AB$ by $O$ such that $PAD\cong OBC$. Then $\angle PDA = 20°$, but $\angle POA = \angle OAB = 20 °$, so $AODP$ is cyclic. But then $\angle CBO = \angle DAP = \angle DOP = \angle CDO = 40°$.
Build a parallelogram $ABDE$. Then $$\angle DEA = \angle ABD = 135^\circ = \angle DCA,$$ hence $ADEC$ is cyclic. Moreover, $\angle EAD = \angle BDA = x$, hence $$\angle CAE = \angle CAD - \angle EAD = 2x-x=x.$$ As a consequence, $CE=DE$ because the angles subtended by arcs $CE$, $DE$ of the red circle are equal.
Now, it is easy to calculate that angle $CBD$ equals $90^\circ + x$ and that the concave angle $CED$ equals $180^\circ + 2x$. Since $CE=DE$, it follows that $B$ lies on the circle with center $E$ and radius $CE=DE$. Hence $\angle DEB = 2\angle DCB = 90^\circ$. Since $DE=BE$, the right triangle $DEB$ is isosceles.
Let $F$ be the midpoint of $BE$. Let $G$ be the midpoint of $BD$ and $H$ be the midpoint of $BG$. Easy to see that triangles $GFB$ and $GFH$ are isosceles right triangles, hence $HF = HB = \frac 12BG = \frac 14 BD$, so $HD = BD - BH = 4HF - HF = 3HF$. This yields $\tan x = \frac{HF}{HD} = \frac 13$ so the answer is $$x = \arctan \frac 13.$$
Best Answer
I don't offhand see how the problem can be solved using your auxiliary lines. Note it's generally better to first check adding lines within a geometric object before trying to extend and connect lines outside of the object. In particular for your problem, consider joining $D$ to $M$ instead.
Then since $\measuredangle MGC = \measuredangle MGD = 90^{\circ}$, SAS (Side-Angle-Side) at $G$ shows that $\triangle MGC \cong \triangle MGD$. By the properties of a parallelogram, as you've already stated, $\measuredangle MCG = 180^{\circ} - 110^{\circ} = 70^{\circ}$, so $\measuredangle MDG = \measuredangle MCG = 70^{\circ}$. Since the parallelogram properties also give that $\measuredangle CDA = 110^{\circ}$, then
$$\measuredangle MDF = \measuredangle CDA - \measuredangle MDG = 110^{\circ} - 70^{\circ} = 40^{\circ} \tag{1}\label{eq1A}$$
Next, similar to what I did at $G$, using SAS at $F$ shows that $\triangle MFD \cong \triangle MFA$. This gives, along with using \eqref{eq1A}, that
$$\measuredangle MAD = \measuredangle MDF = 40^{\circ} \tag{2}\label{eq2A}$$