Let the bottom edge of a rectangular mirror on a vertical wall be parallel to and h feet above the level floor. If a person with eyes t feet above the floor is standing erect at a distance d feet from the mirror, what is the relationship among h, d and t if the person can just see his own feet in the mirror?
I don't quite understand the solutions here and here.
There's an explanation here:
pretend the mirror is a hole into which you are looking at another object.
1) If we represent this the X,Y axis, with the line x = 0 representing the wall, and x = d representing the person. Then, the reflection of the person is the line x = -d.
2) Then, because he can just see his feet in the mirror: draw a line from the top of the person's head to the feet of his reflection. Imagine if another person is actually standing there and you can only see down to just his feet through the "hole".
3) This creates two congruent triangles. Giving the result t = 2h, regardless of d.
I don't quite understand the last step. Please help me visualise this:
Let $d > 0$. The person stands from $(d,0)$ to $(d,t)$. The reflection stands from $(-d,0)$ to $(-d,t)$. The bottom edge of the mirror is at $(0,h)$ where the top edge is at $(0,h_1)$ where $h_1 > h$.
So the feet are in $(d,h)$ to $(d,h_1)$?
Then we're supposed to arrive at
$$\frac{t}{h} = \frac{2d}{d} \to t = 2h$$
I'm so confused. Please help.
Here's a picture:
Here, the person is standing at $B$, with his/her head at $D$. The mirror's bottom is at $C$, which is $h$ feet above the floor. Finally, the "feet of the reflection" are at point $A$.
Triangles $\triangle ABD$ and $\triangle AEC$ are similar, so that $\frac{t}{h} = \frac{2d}{d}$.
BCLC edit: $(0,h)$ is on the line $AD$ which is given by $$y=\frac{t}{2d}(x+d)$$ hence $$h=\frac{t}{2d}(0+d) \to h=\frac{t}{2}$$