Parallel Lines 030D0A

1. **Problem statement:** Given triangle $ABC$ with $\angle ABC = 70^\circ$, $\angle ACB = 65^\circ$, and a line $MN$ passing through points $M$ on $AB$ and $N$ on $AC$ such that $\angle AMx = 110^\circ$ (where $Mx$ is the horizontal line through $M$ and $N$). 2. **Find:** a) Prove that $MN \parallel BC$. b) Find $\angle ANM$. --- 3. **Step a) Prove $MN \parallel BC$:** - First, find $\angle BAC$ in triangle $ABC$ using the triangle angle sum: $$\angle BAC = 180^\circ - \angle ABC - \angle ACB = 180^\circ - 70^\circ - 65^\circ = 45^\circ.$$ - Note that $\angle AMx = 110^\circ$ is an exterior angle at $M$ formed by line $MN$ and segment $MA$. - Since $Mx$ is horizontal and $\angle AMx = 110^\circ$, the angle between $MN$ and $AB$ at $M$ is $110^\circ$. - The interior angle at $M$ between $MN$ and $AB$ is then: $$180^\circ - 110^\circ = 70^\circ.$$ - Because $M$ lies on $AB$, the angle between $MN$ and $AB$ at $M$ is $70^\circ$. - Similarly, consider $\angle ABC = 70^\circ$. - Since $MN$ passes through $M$ and $N$ on $AB$ and $AC$ respectively, and the angle between $MN$ and $AB$ at $M$ equals $\angle ABC$, by the Alternate Interior Angles Theorem, $MN \parallel BC$. --- 4. **Step b) Find $\angle ANM$:** - Since $MN \parallel BC$, angles $\angle ANM$ and $\angle ACB$ are corresponding angles. - Given $\angle ACB = 65^\circ$, so: $$\angle ANM = 65^\circ.$$ --- **Final answers:** - a) $MN \parallel BC$. - b) $\angle ANM = 65^\circ$.