Show that if dist(P, ℓ1) = dist(P, ℓ2) then P is on the bisector of the angle

Question

In the image above line ℓ1 refers to $AQ$ and the ℓ2 refers to $AR$. It is given that dist(P, ℓ1) = dist(P, ℓ2). I think I have to show that $PA$ is the bisector of $\angle A$ but I don't know where to start, please any help appreciated.

Answer 1

What you mean by dist(P, ℓ1) = dist(P, ℓ2), is I guess the perpendicular or the shortest distance of ℓ1 and ℓ2 from $P$. So that is equivalent to the condition $PQ = PR$. So in triangles $AQP$ and $APR$:

• $\angle AQP= \angle ARP = 90^{\circ}$
• $AP$ is common.
• $PQ = PR$.

Hence, triangles $AQP$ and $ARP$ are congruent by the RHS criterion. So, $\angle PAQ=\angle PAR$ and hence $P$ lies on the bisector of $\angle A$.

Answer 2

From your hypothesis, $QP=RP$. So, from the Pythagorean Theorem, the triangles $AQP$ and $ARP$ are equal, since $AP$ is common, and $$AQ^2=AP^2-QP^2=AP^2-RP^2=AR^2\Rightarrow AQ=AR.$$ Therefore the angles on the left are equal, so $P$ lies on the bisector.