I would like to know about a software that will help me show construction steps to the students with using a Compass/Straight Edge/Protractor/Divider. Here is example video below.
[Math] Dynamic Geometry Software for teach construction
educationgeometric-constructiongeometry
Related Solutions
Given $A,B$. Find their midpoint $M$:
Find a line $\ell\|AB$ (see below). On $\ell$, mark points $P,Q,R$ with $|PQ|=|QR|=u$. Let $PB$ and $RA$ intersect in $Z$. Then $ZQ$ intersects $AB$ in $M$.
Given a line $\ell_1$, find a distict line parallel to it:
On $\ell_1$, find $C,D$ with $|CD|=u$. Draw a line $\ell_2\ne\ell_1$ through $C$. Find $E\ne C$ on $\ell_2$ with $|DE|=u$. Find $F\ne C$ on $\ell_2$ with $|FE|=u$. Find $G\ne E$ on $\ell_2$ with $|GF|=u$. Then $FG\|\ell_1$.
I got introduced in such stuff by: P.Schreiber, Theorie der geometrischen Konstruktionen, Berlin 1975. I'm sure there's more readable and modern literature available.
One important step is always to make clear which construction steps are available with a given set of instruments. Here we use:
- Given two points $A,B$, find the line $AB$ through them
- Given two non-parallel lines $\ell_1,\ell_2$, find their point of intersection.
- Given a point $A$ and a line $\ell$ less that $u$ apart, find the two points $P,Q$ on $\ell$ with $|AP|=|AQ|=u$.
Often one needs to pick random elements (and then show that the finalk result does not depend on the random choices):
- Pick a "random" point in the plane
- Pick a "random" line through a point $A$ (can be reduced to picking a random point $B$ and finding $AB$)
- Pick a "random" point on a line $\ell$ (can be reduced to picking a random point in the plane, then a random line through that point, then intersect this with the given line)
In the construction above, we are given the circle centred on $O$ (black) and the point $P$. We pick an arbitrary point $A$ on the circle and construct two more circles (in light grey) which intersect in $B$ and $C$; a segment of the tangent is then $PC$. $Z$ is a point on the black circle, on the other side of the chord $PB$ from $A$, and is solely a component of our proof: that $\angle OPC=90^\circ$. Line segments of the same length have the same colour (except the ones incident to $Z$).
Say $\angle BPA=x$. Since $\triangle PAB$ and $\triangle CAP$ are congruent and isosceles, $\angle APC=x$.
$\angle PAB$ is $180^\circ-2x$, and since quadrilateral $PABZ$ is cyclic, $\angle PAB+\angle BZP=180^\circ$, or $\angle BZP=2x$. By the inscribed angle theorem, $\angle BOP=4x$.
Since $\triangle BOP$ is isosceles, $\angle OPB=\frac12(180^\circ-4x)=90^\circ-2x$. Then $$\angle OPC=\angle OPB+\angle BPA+\angle APC=90^\circ-2x+x+x=90^\circ$$ Hence $PC$ is tangent to the black circle at $P$.
Best Answer
Tabula looks like it might fit you requirements, but I only briefly looked at the screenshots:
I found it after scanning this list for "extras". Maybe you'll find more there.