25. The diagram shows a triangle $F H I$, and a point $G$ on $F H$ such that $G H=F I$. The points $M$ and $N$ are the midpoints of $F G$ and $H I$ respectively. Angle $N M H=\alpha^{\circ}$. Which of the following gives an expression for $\angle I F H$ ?

A $2 \alpha^{\circ}$
B $(90-\alpha)^{\circ}$
C $45+\alpha^{\circ}$
D $\left(90-\frac{1}{2} \alpha\right)^{\circ}$
E $60^{\circ}$
so $P G=\frac{1}{2} I G$. Since $M G=\frac{1}{2} F G$, the triangle $P M G$ is similar to $I F G$, and in particular, $P M=\frac{1}{2} I F$. However, we know $I F$ is equal in length to $G H$ so we have
$P N=\frac{1}{2} G H=\frac{1}{2} I F=P M$, so triangle $M N P$ is isosceles and $\angle P M N=\angle P N M=\alpha$.
Since triangles $P M G$ and $I F G$ are similar, we have $\angle I F G=\angle P M G=\alpha+\alpha=2 \alpha$.
25. A We start by drawing the line segment $I G$. Let $P$ be the point on $I G$ such that $P N$ is parallel to $F H$.
The angle $P N M$ is alternate to $N M H$ so
$\angle P N M=\alpha$. Also, the triangle $P N I$ is similar to the triangle $G H I$ (the angles of each triangle are clearly the same); moreover since $N$ is the midpoint of $H I, P N=\frac{1}{2} G H$. Also $I P=\frac{1}{2} I G$,