Is the law of reflection against math? In my textbook of science (class 10), it is given that for any mirror, the angle of incidence is equal to the angle of reflection.
Here I m talking about spherical mirror. A convex mirror.
All rays which are incident on a mirror meet at a point called focus. And it is also given that focus is the midpoint of radius of the sphere.
The below is the diagram given in the book. It shows that if the object is at a known distance from the convex mirror, it will form this type of image. It shows that if a light ray is incident parallel to the principal axis, it will passes through focus.

I tried to calculate this:-
$$\angle{GDH} = \angle{HDI} = \alpha$$
$$PF = FC = x$$
$$\angle{FDC} = \angle{GDH} = \alpha \text{(Vertically Opposite Angles)}$$
$$\angle{DCF} = \angle{HDI} = \alpha$$
$\triangle{FCI} $ becomes isosceles.
$$\therefore DF = FC = x$$
Join DP
$\triangle{FDP}$ also becomes isosceles as FD = PF = $x$.
$$\angle{PDF} = \angle{DPF}$$
But in $\triangle{CDP}$
$$\angle{DPF} = \angle{PDC}$$
This means $\angle{FDC} = 0$ and F = C. But it is given that $2PF = PC.$
This contradiction has occured because of our assumption that $\angle{GDH} = \angle{HDI}$. So $\angle{GDH} ≠ \angle{HDI}$. But the law of reflection says that these angles are equal.
Can anyone please help me with this? I think I m wrong somewhere but I can't find it out.

 A: The geometrical optics typically taught in year 10 science uses a number of simplifying approximations.
The correct shape for a mirror which focuses parallel light rays to a single point is a parabola, which would have its axis parallel to the rays. A spherical mirror would reflect rays from the centre of the sphere back to the same point (the centre). If the curvature of the mirror is small a spherical mirror is a fairly good approximation to a parabolic mirror. The approximating sphere has its centre at twice the focal length from the mirror.
You have correctly identified that the spherical mirror does not behave in the same way as the parabolic mirror. The two are reasonably close because the equation for a semicircle, $y=\sqrt{1-x^2}$ can be expanded to an infinite series $y=1-\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^6}{16}-\frac{5x^8}{32} +$ more terms. When $x$ is small this is fairly close to the quadratic function $y=1-\frac{x^2}{2}$, which describes a parabola.
The simplifying approximation used in high school science classes often assumes that the mirror is flat enough for reflection to occur on the plane of the mirror. This is a simplifying approximation, which works well enough for many purposes. However it is only an approximation.
A: As the angle of incidence equals the angle of reflection, we have that $\triangle DFC$ is isosceles - this means that $|FC|=|FD|$
This in turn means that $F$ lies on the intersection of the perpendicular bisector of $CD$ and $PC$, which does not give $|PF|=|FC|$.
