Let $\angle A$ be given with two sides $l_1,l_2$, and a point $K$ in the interior of the angle. How could I construct two points $p_1,p_2$ on $l_1,l_2$ respectively, so that the midpoint of $p_1$ and $p_2$ is exactly $K$?

I don't know where to start. I thought of constructing a perpendicular bisector of $AK$, but that has nothing to do with the two sides of the angle. Then I thought of constructing the angle bisector of $A$, but it need not pass $K$. Any suggestions?


Suppose point $A' \neq A$ on line $AK$ satisfies $A'K = AK$, and line $l_1'$ and $l_2'$ passing through $A'$ are such that $l_1'$ is parallel to $l_1$ and $l_2'$ is parallel to $l_2$.

Now suppose $P_1 = l_1 \cap l_2'$ and $P_2 = l_2 \cap l_1'$, then $AP_1A'P_2$ is a parallelogram. Thus $K = AA' \cap P_1P_2$ is the midpoint of $P_1P_2$.



Let$ l_1,l_2, $ be intersecting lines at point $A$ with an acute angle $\alpha.$

Let $K$ be any point within.

For definiteness consider the $1$st quadrant, $l_1$ the 'lower' line .

1)Draw a parallel through $K$ to line $l_1$, intersecting line $l_2$ at $B.$

2) Draw a circle with center $B$ and radius = length $AB$, intersecting $l_2$ at point $P_2.$

3) The line $P_2K$ intersects $l_1$ at $P_1.$


Consider the intersecting lines $P_2A$$(l_2)$and $P_2P_1$.

1)Line $KB$ and line $l_1$ $(AP_1)$ are parallel.

Intercept Theorem :

$P_2B:BA= P_2K:KP_1=1:1.$

Comments welcome.


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