In an isosceles triangle $ABC$ show that $PM+PN$ does not depend on the position of the chosen point P.

Question

Let $ABC$ be a triangle such that $AB = AC$. Let $P$ be a point in $BC$. Let $M, N$ be the feet of the perpendiculars from $P$ to $AB$ and $AC$ respectively. Show that the value of the sum $PM+PN$ does not depend on the position of the chosen point P.

My try

I didn't progress a lot in this problem but i'm going to put what i saw:

First, i tried to draw the line $AP$, in that way the right triangles $AMP$ and $ANP$ have the same hypotenuse. After that i tried Pythagoras, but i found nothing.

After that i realized that the right triangles $MBP$ and $NCP$ are similar. I tried some operations with the properties of similar triangles, but also i found nothing.

Any hints?

Answer 1

Observe that $$\frac{AB·PM}{2}+\frac{AC·PN}{2}=\frac{BC·h_a}{2}=\text{Area of }\Delta ABC$$ Since $AB=AC$

$$PM+PN=\frac{BC·h_a}{AB}=h_b=h_c$$

which is constant (and therefore doesn't depend on $P$'s position). $\;\blacksquare$

Answer 2

Hint : let $N'$ be the point symmetrical of $N$ with respect to side $BC$. Recognizing that $M,P,N'$ are aligned (use angles), your issue is equivalent to show that distance $MN'$ remains the same when $P$ is varying on $BC$.

Answer 3

$PBM$ is similar to $PCN.$

$MP$ is proportional to $BP$ as $NP$ is to $CP$ as $MP + NP$ is proportional to $BP + CP$

$BP+CP = BC$ and does not vary based on the location of $P.$

Answer 4

Hint: Draw in the altitudes from $B$ and $C$. From there? Either work with similar triangles, or draw in more things. It's frequently a good idea to add to a diagram in a geometry problem.

Also, a picture: