Isosceles triangle -As vertex(A) approaches midpoint M of BC,what happens to point(P) which intersects side AC.

Question

Please help me with question (a).

Here I'm trying to prove the property that $\overline PM$ $\backslash$ $\backslash$ $\overline AB$, then use the property of slope to answer.But I failed to prove that $\overline PM$ $\backslash$ $\backslash$ $\overline AB$,therefore I'm looking for the alternative method to deal with this question.

figure for question

Answer 1

Let $x$ be the length of the altitude, $\alpha$ is the base angle. We can also assume without loss of generality that the base length is 1 unit. Then, from $\triangle{BPC}$ we have $$\frac{PC}{\sin \alpha/2}=\frac{BC}{\sin (\pi-\alpha-\alpha/2)}$$ or $$PC=\frac{\sin \frac{\alpha}{2}}{\sin \frac{3\alpha}{2}}=\frac{1}{-4\sin^2 \frac{\alpha}{2}+3}=\frac{1}{1+2\cos \alpha}$$ You can see what the limit will be when $\alpha$ goes to zero. If you want to graph it, however, you can use that $\tan \alpha=2x$ and $\cos \alpha=\frac{1}{\sqrt{1+\tan^2 \alpha}}$. Can you take it from here?

Answer 2

This isn't really an improvement on Vasya's answer, but I've thought about this too long to let it go. So ...

The law of sines from trigonometry tells us

$ \displaystyle \frac{\sin B/2}{AP} = \frac{\sin B}{BP} $ and $ \displaystyle \frac{\sin{A}}{BC} = \frac{\sin B}{AC} $

So we can eliminate $\sin A$ and get

$ \displaystyle \frac{BP}{AP}=\frac{BC}{AC}\frac{\sin B}{\sin B/2} $

Now, A is heading for the midpoint of BC, so when it reaches it $BC/AC$ will equal 2. At the same time $\sin B$ and $\sin B/2$ are both heading to zero. But you should be able to use calculus to show that in this limit $\sin B / \sin B/2 \rightarrow 2$. From here, finding the location of $P$ is all algebra.