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I have geometrical question which I'm trying to solve for a while now and it goes like this :

given a Line AB and 2 arbitrary points on one side of that line (C and D).
suggest a method to find a point M along AB so that ∠CMA = ∠DMB

enter image description here

can I please get direction to the solution ? Thanks!

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  • $\begingroup$ Did you try using coordinate geometry?,You have equation of line and 2 points,since angle subtended is equal,you can equate the slopes and find the point $\endgroup$ Jul 20, 2016 at 8:01
  • $\begingroup$ this is a valid approach in case the coordinates are given , in this case the question is about solving it using geometrical properties , more of using compasses and a ruler kind of a solution rather than analytic geometry one. $\endgroup$ Jul 20, 2016 at 8:27

2 Answers 2

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Hint: Having $\angle AMC = DMB$ is equivalent to having the perpendicular to $AB$ through $M$ to bisect $\angle CMD$. Recall that a light ray reflecting off a mirror has exactly that property, which means that $M$ is where $C$ needs to look at to see $D$ in the mirror $AB$. Thus you can construct $M$ by constructing $D'$ to be the reflection of $D$ in $AB$ and then $M = AB \cap CD'$.

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  • $\begingroup$ this solution is so elegant! how didn't I think of it :) Thank you so much!! $\endgroup$ Jul 20, 2016 at 11:52
  • $\begingroup$ @DimaShifrin: You're welcome! And thanks for the appreciation! =) $\endgroup$
    – user21820
    Jul 20, 2016 at 12:58
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Here is one way. From $C$ and $D$, draw perpendicular lines to the line $AB$. So you get $CC_1$ and $DD_1$.

Now, if you find a point $M$ on $AB$ that satisfies the following equation, then you have the required point.

$\frac{C_1M}{D_2M}=\frac{C_1C}{D_1D}$

It is easy to see that triangles $CC_1M$ and $DD_1M$ are similar.

Now it remains to find a way to find edges that their length add up to the length of $C_1D_1$ and have the required fraction. An easy way is to draw an edge with length $|CC_1|+|DD_1|$ from $C_1$ (with an arbitrary angle, such that the angle between the edge and $AB$ is less than $90$ degrees). The end point is called $X$. make edges $CC_1$ and $DD_1$ on $C_1X$ and call the point that they meet $Y$. Then connect $X$ to $D_1$. Draw a line parallel to $D_1X$ from $Y$ to intersect the line $AB$. the intersection is $M$.

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  • $\begingroup$ Thank you for the solution! $\endgroup$ Jul 20, 2016 at 11:58
  • $\begingroup$ Hope it it helpful. $\endgroup$
    – Med
    Jul 20, 2016 at 12:47

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