Valentin Vornicu wrote: We denote by $Q$ the intersection between the line $AB$ and the side $AB$... That'd just be AB, right?

nr1337 wrote: Valentin Vornicu wrote: We denote by $Q$ the intersection between the line $AB$ and the side $AB$... That'd just be AB, right? It's $CU$ actually. I've modified the statement. Thanks for noticing.

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Since the triangles $\triangle QCP, \triangle MCN$ have the same altitudes from the vertices $Q, M$ to the sides $CP, CN$, respectively, equal to the distance between the opposite sides of the rhombus $ABCD$, it is necessary to show that $CP = CN$. Using a parallel projection, we can project the rhombus $ABCD$ into a square $A'B'C'D'$, while preserving all area ratios. Hence, it is sufficient to prove the proposition for a square $ABCD$. Let $a = AB$ be the side of the square and $m = \frac{AM}{AD} = \frac{AM}{a}$, $n = \frac{BN}{BC} = \frac{BN}{a}$, $p = \frac{DP}{DC} = \frac{DP}{a}$. We have to show that $n = p$ regardless of $m$. From similarity of the triangles $\triangle BNU \sim \triangle DMU$, we have $\frac{BU}{DU} = \frac{BN}{DM} = \frac{n}{m - 1}$. From similarity of the triangles $\triangle BQU \sim \triangle DQU$, we have $\frac{BQ}{DC} = \frac{BU}{DU} = \frac{n}{m - 1}$. From similarity of the triangles $\triangle BCT \sim \triangle DMT$, we have $\frac{BT}{DT} = \frac{BC}{DM} = \frac{1}{m - 1}$. From similarity of the triangles $\triangle BTQ \sim \triangle DTP$, we have $\frac{BQ}{DP} = \frac{BT}{DT} = \frac{1}{m - 1}$. On the other hand, $\frac{BQ}{DP} = \frac{BQ}{DC} \cdot \frac{DC}{DP} = \frac{n}{m - 1} \cdot \frac 1 p$. Thus $\frac{1}{1 - m} = \frac{n}{m - 1} \cdot \frac 1 p$ or $n = p$.

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you can prove it using just simillarities in the same way, without complicated stuff like parallel projections, which btw i have no idea what they are(no need to assume abcd = square)