In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$ (AMC 12 2018)
I'm trying to understand all of the solutions but I am stuck at solution 4
In Solution (4)
"Let the mid-point of $\overline{AT}$ be $B$ and the mid-point of $\overline{GT}$ be $C$. Since $\overline{BC}=\overline{CG}-\overline{BG}$ and $\overline{CG}=\overline{AB}-\frac{1}{2}$, we can conclude that $\overline{BC}=\frac{1}{2}$. Similarly, we can conclude that $\overline{BM}-\overline{CN}=\frac{1}{2}$. Construct $ND//BC$ and intersects $\overline{BM}$ at $D$, which gives $\overline{MD}=\overline{DN}=\frac{1}{2}$. Since $\angle{ABD}=\angle{BDN}$, $\overline{MD}=\overline{DN}$, we can find the value of $\angle{DMN}$, which is equal to $\frac{1}{2}T=44^{\circ}$. Since $BM//PT$, which means $\angle{DMN}+\angle{MNP}+\angle{P}=180^{\circ}$, we can infer that $\angle{MNP}=100^{\circ}$. As we are required to give the acute angle formed, the final answer would be $80^{\circ}$, which is $\boxed{\textbf{(E)}}$. (Surefire2019)"
I do not understand $\overline{BM}-\overline{CN}=\frac{1}{2}$ and
$\angle{ABD}=\angle{BDN}$, $\overline{MD}=\overline{DN}$ in which they did not elavorate.
Best Answer
Using the given data, we can write, $$ \begin{matrix} TG=TA-AG=TA-1 & \mathrm{and} & TU=TP-PU=TP-1 \\ \end{matrix} .$$
Since $B$ and $M$ are the midpoints of $AT$ and $AP$ respectively, $BM$ is parallel to $TP$. Similarly, $CN$ is parallel to $TP$, because $N$ and $C$ are the midpoints of $UG$ and $TG$ respectively. Therefore, we also aware that, $$ \begin{matrix} BM=\frac{TP}{2} & \mathrm{and} & CN=\frac{TU}{2}=\frac{TP}{2}-\frac{1}{2} \\ \end{matrix} .$$ Now, we can write, $$BM-CN=\frac{TP}{2}-\left(\frac{TP}{2}-\frac{1}{2}\right)=\frac{1}{2}\tag{1}$$
We also know, that, $$ \begin{matrix} AB=\frac{AT}{2} & \mathrm{and} & GC=\frac{GT}{2}=\frac{AT}{2}-\frac{1}{2}=AB-\frac{1}{2} \\ \end{matrix} .$$ Since $BG=AB-AG=AB-1$, we have, $$BC=CG-BG=\frac{1}{2}.$$
Since $ND//BC$ and $NC//DB$, we have $$ \begin{matrix} BD=CN & \mathrm{and} & DN=BC=\frac{1}{2} \\ \end{matrix} .$$
To find $DM$, we write, $$MD=BM-BD=\left(CN+\frac{1}{2}\right)-CN=\frac{1}{2}=DN.\tag{2}$$
Since $ND//BC$, $\measuredangle ABD$ and $\measuredangle BDN$ are alternate angles. Therefore, they are equal, i.e. $$\measuredangle ABD=\measuredangle BDN \tag{3}.$$
Since $MD=DN$, $MDN$ is an isosceles triangle. Therefore, $$\measuredangle DMN=\frac{180^0-\measuredangle ABD}{2}=\frac{180^0-\measuredangle ATP}{2}=\frac{180^0-56^0 - 36^0}{2}=44^0.$$
Finally, we have, $$\measuredangle NMA=\measuredangle NMD + \measuredangle BMA = \measuredangle NMD + \measuredangle TPA=44^0+36^0=80^0.\tag{4}$$