The given figure is a square and $M$ is the midpoint of $AB$. traingle $AMN$ and $BMP$ are congruent also traingle $NMC$ congruent $CMP$, The angle's that are in red and orange are equal.
If the length of one side of the square $ABCD$ is 4 cm , then, using equiangular triangles , show that $AN=1$ cm
Any suggestions on the equal angles?
On
First of all triangles $NMC$ and $CMB$ are not congruent. They are similar. Congruent are $NMC$ and $CMP$. So I am not quite sure what initial information you are given, but let's say you are told that
Version 1. $M$ is the midpoint of $AB$ and that points $N$ and $P$ are chosen on lines $AD$ and $CB$ so that line $NP$ passes through the midpoint $M$ and $CN = CP$. triangles $NMC$ and $CMP$ are congruent.
Version 2. $M$ is the midpoint of $AB$ and line $NP$ is chosen so that it goes through $M$ and is orthogonal to line $CM$.
In both versions, you show that first that triangle $AMN$ is congruent to triangle $BMP$. That allows you to conclude that $NM = PM$ and using that you can show that triangles $NMC$ and $CMP$ are congruent. Next, you see that $\angle \, AMN = \angle \, BMP = \alpha$ (red angles) and $\angle \, MCN = \angle \, MCP = \beta$ (orange angles).
Hint 1. If $NMC$ and $CMP$ are congruent and they share a common edge $CM$, then how much is $\angle \, CMP\, \, $?
Hint 2. How much is angle $\angle \, P$ as an angle in triangle $MBP$ and in triangle $CPM$. Form here what is the relationship between $\alpha$ and $\beta$?
Hint 3. What can you say about the angles of triangles $CMP$ and $MBP$? Consequently what kind of triangles are these and what is the relationship between the lengths of their edges?
HINT.
Triangles $MBC$ and $MAN$ are equiangular and thus similar because...