Geometric construction of a triangle given 2 lines and a point

Question

Lines $t$ and $s$ and point $P$ are given. Line $t$ is a perpendicular from the centroid of triangle $ABC$ to $BC$, line $s$ is a bisector of angle $\angle ABC$, and $P$ is the midpoint of $BC$. Construct triangle $ABC$.

I constructed a perpendicular from $P$ to $t$ and its intersection with the line s is point $B$. Point $C$ is the centrosymmetric image of point $B$ with respect to $P$, and point $A$ lies on the line $BP'$, where $P'$ is the axisymmetric image of point $P$ with respect to $s$. How do I find point $A$?

Answer 1

Note. the line and point are renamed in figure.

You can also use this fact that the distance between mid point of base D and bisector of angle KAB is equal to the radius of nine point circle.Draw a perpendicular f from mid point D on line t passing the centeroid of triangle, base AB is coincident with this line, A is intersection of this line with bisector of angle KAB. Also Use this fact that the radius of circumcircle is double that of nine point circle and that the center G of this circle is on a perpendicular on AB at mid point D. Draw circumcircle , it intersects line f on A and B . Take an arbitrary point like H on base AB as the foot of altitude from vertex K. Draw perpendicular k on AB from H.Draw perpendicular bisector n of DH. Draw a circle tangent on bisector of angle KAB(line s in your statement), it intersects n at N. N is the center of nine point circle, draw this circle. Find mirror of G about N, it will be G' on k, this is orthocenter ( where altitudes intersect) of triangle.Connect A and B to G' and extend. Draw perpendicular from A and B to these lines. These perpendiculars intersect at K which is the third vertex of triangle. Since H is arbitrary on AB, there can be numerous triangle with given parameters in statement.

Answer 2

Construct the unique line that is perpendicular to $t$ and passes through $P$. Note that $t$ is perpendicular to $BC$ and $P$ lies on $BC$. It follows that the line in consideration contains the line segment $BC$. Call this line as $l$.

However, the line $s$ is the angle bisector of $\angle ABC$. In particular, $B$ lies on $s$. It follows that $B$ lies on both $l$ and $s$. Thus, $B = l \cap s$, the intersection of the two lines.

Since $P$ is the midpoint of $BC$ and $B,P$ lie on $l$, it follows that $C$ also lies on $l$. Therefore, $C$ is the reflection of $B$ about the point $P$.

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We will now construct the point $A$. Let us set a coordinate system in place for this. Suppose that $B$ is the origin of the system, with coordinates $B=(0,0)$. We let the $x$-axis of the system be the line $l$, oriented in such a way that $C = (c,0)$ where $c>0$. The $y$-axis is then the line which is perpendicular to $l$ (alternately, parallel to $t$) and passes through $B$. We choose the orientation of the $y$-axis in such a way that the line $s$, which is the bisector of $\angle ABC$, passes through the first quadrant.

Attached below is a picture with the value $c=2$. Note that the point $D = t \cap l$ is available for us : in this case, we have $D = (0.6,0)$. In general , $D = (d,0)$ for some $d$.

Since $P$ is the midpoint of $BC$, $P = (\frac c2,0)$. (In the case above, $P = (1,0)$).

Suppose that the point $A = (y,z)$. Then, observe that the centroid $O$ of the triangle ABC has the coordinates $(\frac{c+y}{3},\frac z3)$. However, the line $t$ contains the centroid $O$, therefore the centroid and the point $D$ have the same $x$ coordinate, which equals $d$. In other words, $$ d = \frac{c+y}{3} \implies y = 3d-c. $$ To find $z$, we observe that $s$ is the angle bisector of $ABC$. Therefore, we merely replicate the angle between $l$ and $s$ on the other side of $s$ to obtain a line $l'$ such that the angle between $l$ and $s$ at $B$ equals the angle between $l'$ and $s$ at $B$. The point $A$ must lie on $l'$, which can be easily constructed.

$A$ must also lie on the line $y = 3d-c = 3BD -BC$. Therefore, we first find the point $E = (3d-c,0)$. To do this, first measure $3BD-BC$ (all these distances are available) : if it's positive, then $E$ is the point in $l$ which is distance $y$ to the right of $B$ in this coordinate system, otherwise $E$ is the point in $l$ which is at distance $y$ to the left of $B$. Let $j$ be the line perpendicular to $l$ which passes through $E$.

Then, $A = j \cap l'$.

To prove that $ABC$ is the desired triangle, we observe that indeed $s$ is the angle bisector of $ABC$ by construction (since $A$ lies on $l'$ and $C$ lies on $l$, and $l'$ is chosen to be such that $\angle l's = \angle ls$ at the point $B$).

Of course, $P$ is the midpoint of $BC$ as well, because we found $B$ and then created $C$ so that $P=BC$.

That $t$ is perpendicular to $BC$ follows because $B,C$ lie on $l$ which is perpendicular to $t$ by choice. That $t$ contains the centroid $O$ of $ABC$ follows by the following argument : the median is always divided by the centroid in the ratio $2:1$. That is, $AO:OP = 2:1$. In particular, by the section formula, this ratio is retained by the coordinates. So the $x$ coordinate of $O$ equals $\frac{2}{3}$ times the $x$ coordinate of $A$ plus $\frac 13$ times the $x$ coordinate of $P$, which is $\frac{2}{3}y + \frac{1}{3}\frac c2 = d$ since $y=3d-c$. The line $t$ is defined by the equation $x=d$ given our coordinate system, so $O$ lies on $t$, as desired.

To show the complete image :

Here, $d = 0.6$ and $c=2$, so $3d-c = 0.2$. That is, $E = (-0.2,0)$ is chosen to the left of $B$. Once the line $l'$ is constructed as mentioned, we drew the perpendicular to $l'$ at $E$ which is $j$, and $A = j \cap l'$. I've just shown the point $O$ which is the centroid of $ABC$ in the picture, to confirm that it lies on $t$.

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Here's another example of the same situation. We were given $P,t,s$, and draw the line $l$ passing through $P$ and perpendicular to $t$. Then $B = l \cap s$ and $C$ is the point $B$ reflected about the point $P$. We ensure that the coordinate system is centered at $B$, the line $s$ passes through the first quadrant, the $x$ axis is $l$, $C = (7,0)$ (so $c=7$), and $l \cap t = D = (3,0)$, hence $d = 3$.

We have the $y$ coordinate of $A$ which equals $3d-c = 9-7=2$ (in this case, it's positive). Then we reflect the angle between $l$ and $s$ onto the other side of $s$ to get a new line $l'$, and find the point $E = (2,0)$. Finally, the perpendicular to $l$ at $E$, denoted by $j$, intersects $l'$ at the point $A$, as desired. Once again, we show that $O$ lies on $t$ for completeness.

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If one wishes to make the construction coordinate free, that's not really a problem since the axes are already represented by the lines $t$ and $l$. So let me write down, for completeness, a coordinate-free version of the construction above.

Given $P,t,s$,

• Let $l$ be the unique line that is perpendicular to $t$ and passes through $P$.

• $B = l \cap s$.

• $C$ is the reflection of the point $B$ about the point $P$.

• Find the point $D = BC \cap t$.

• Find the quantity $3BD-BC$ (no need for a ruler, since these quantities are all available on paper for us to use). Let $E$ be the point on the line segment $l$ such that $BE$ is equal to the signed distance $3BD-BC$, where we assume that $BC$ is a positively signed distance. In other words, if $3BD-BC>0$ then $E$ and $C$ are on the same side of $B$, otherwise they're on different sides of $B$.

• Let $\alpha$ be the angle formed between the lines $s$ and $l$ at the point $B$. Reflect $\alpha$ about the line $s$ to get a new line $l'$.

• Draw a perpendicular to $l$ at the point $E$, which is the line $j$.

• $A = j \cap l'$.

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If one wishes to get a coordinate-free proof of this construction as well, then here is a sketch, where we skim over some orientation issues that didn't trouble us in the coordinate system scenario.

We know that $A$ must certainly lie on $l'$. Drop a perpendicular from $A$ to $l$ and let $E$ be the point on $l$ which is hit by it. Now, look at the triangles $AOP$ and $EDP$. They share a common angle at $P$ and both contain a right angle. Hence, they're similar to each other. As a consequence, $AO:OP = ED:DP$. However, $AO:OP = 2:1$ by the property that the centroid divides every median in that ratio, therefore $ED:DP = 2:1$. Now, some simple vector algebra yields the result that $BE = 3BD-BC$ as signed distances, hence $E$ is rightly located by our algorithm.

Answer 3

$\mathrm{Fig.\space 1}$ shows what you have done up to now. The given information is shown in $\color{red}{\text{red}}$. The black lines and points are what you have constructed. By the way, we renamed your axisymmetric image of point $P$ as $Q$.

$\mathrm{Fig.\space 2}$ shows what you should do to locate the vertex $A$ of $\triangle ABC$. Mark two arbitrary points $M$ and $N$ on the given line $t$ and draw the two rays $PM$ and $PN$. Now locate the point $K$ on $PM$ such that $MK=2PM$. Similarly, mark the point $L$ on $PN$ such that $LN=2PN$. To complete the construction join the points $K$ and $L$ to intersect the line $BQ$ as shown in the diagram. The point of intersection of $BQ$ and $KL$ is the sought vertex $A$.

I think that you are versed in geometry to figure out why the steps described above lead to what you were looking for.

Answer 4

What one should consider is the fact that the centroid of a triangle divides a median at a ratio 1:2.

The picture below depicts the situation in the problem in kaykay's posting, that is, $P$ is the middle point of segment BC, line $t$ contains the centroid of $\triangle ABC$ and is perpendicular to BC, and line $s$ is a bisector of the angle at $\angle ABC$. $A$ is to be determined. The line trough $B$ and $P'$ is such that the $\angle CBP'$ is twice the angle form between line $s$ and segment BC.

Denote by Q the point at which the line t intersects the segment BC. Using the fact mention above, all one needs to do is to find a point X on the line going from Q to B so that the length of QX is twice the length of segment PQ. From X draw a perpendicular line call it $\ell$ to the segment BC. $\ell$ will intersect the line that contains $B$ and $P'$ at the desired point A.

This works because of similarity of triangles: being rectangular triangles, $\triangle PXA$ and $\triangle PQF$ (F being the Centroid of $\triangle ABC$) are similar and $$\frac{PF}{PA}=\frac13=\frac{PQ}{PX}$$ Hence $QX=2QP$.

Answer 5

Proceeding from where you were, draw line $r$ from $P$ parallel to $t$.

There are now three cases.

1. $\angle ABC < \pi /2$. Let $Q$ be the intersection of $r$ with line $AB$ and $R$ be the intersection of $t$ with line $AB$. Then, if $T$ is symmetric to $Q$ w.r.t $R$, $A$ is symmetric to $R$ w.r.t $T$. (Thales Theorem on $AP$ and $AB$.)
2. $\angle ABC > \pi /2$. Let $Q$ be the intersection of $r$ with line $AB$ and $R$ be the intersection of $t$ with line $AB$. Then, if $T$ is symmetric to $Q$ w.r.t $R$, $A$ is symmetric to $R$ w.r.t $T$. (Thales Theorem on $AP$ and $AQ$.)
3. $\angle ABC = \pi/2$. Let $Q$ be the intersection of $r$ with line $s$, $R$ the point symmetric of $P$ w.r.t. $Q$, and $T$ the point symmetric of $Q$ w.r.t. $R$. Then we find side $AC$ connecting $C$ with $T$. (Why?)