In the very beginning, I'm going to refer to a post containing key information required in the following proof and especially the answer by @EmilioNovati.
On the edges $\overline{AB}$ and $\overline{AD}$ of the parallelogram $ABCD$ there are points $M$ and $N$ respectively s.t. $MN\parallel BD$. Prove that $\operatorname{Area}(\Delta MBC)=\operatorname{Area}(\Delta NCD).$
My approach:
Let $E$ be the intersection point of $BD$ and $CM$ and let $F$ be the intersection point of $BD$ and $CN$. Let's observe the triangles $\Delta BCE\;\&\;\Delta CDF$ and $\Delta BEM\;\&\;\Delta DFN$.
Let $C_1\in BD$ s.t. $CC_1\perp BD$. Then $\Delta BCE\;\&\;\Delta CDF$ have the common altitude $\overline{CC_1}$.
Let $M_1,N_1\in BD$ s. t. $MM_1\perp BD$ and $NN_1\perp BD.\;\implies MM_1\parallel NN_1.$
$MN\parallel BD\implies |MM_1|=|NN_1|$, so $\Delta BEM\;\&\;\Delta FDN$ have equal altitudes.
Now, the key part:
As @EmilioNovati stated in the answer in the thread, let $G$ be the intersection point of $MN$ and $CD$ and let $B_1$ be the intersection point of $BC$ and $MN$.
$AD\parallel BC\;\&\;MN\parallel BD\implies$ the quadrilateral $B_1BDN$ is a parallelogram and $|B_1B|=|DN|\implies\Delta B_1BM\cong\Delta NDG\implies |B_1M|=|GN|$
According to the Thales intercept theorem:
$$\frac{|BE|}{|B_1M|}=\frac{|DF|}{|GN|}\implies |BE|=|DF|$$
We obtain:
$$\color{red}{\operatorname{Area}(\Delta MBC)}=\operatorname{Area}(\Delta BEM)+\operatorname{Area}(\Delta BCE)=|BE|\frac{|MM_1|+|CC_1|}{2}\\=|DF|\frac{|NN_1|+|CC_1|}{2}=\operatorname{Area}(\Delta FDN)+\operatorname{Area}(\Delta CDF)=\color{red}{\operatorname{Area}(\Delta NCD)}\;\square$$
Picture:
May I ask if there is anything missing or how to improve my proof if necessary? Thank you in advance!
Best Answer
My proof would be
$$\frac{\triangle MBC}{\triangle ABC} = \frac{MB}{AB} = \frac{ND}{AD} = \frac{\triangle NCD}{\triangle ACD}.$$
Since $\triangle ABC = \triangle ACD$, we are done.