0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $P_1P_2$.

$\endgroup$
0
$\begingroup$

Rephrasing:

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 $P_1.$

Reasoning:

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.