# Constructing a right triangle with hypotenuse and bisector

Given two lengths $$\overline{AB} = R$$ and $$\overline{AE} = r$$ with $$R > r$$, how to construct a right triangle $$\triangle ABC$$ with a hypotenuse $$\text{length} = R$$ and the length of the $$\text{bisector} = r$$ as shown below in the figures? Somehow I cannot get the label working for vertex $$C$$, the one on the orange circle, sorry.

The right triangle $$\triangle ABC$$ has the angles $$\alpha + \beta = \pi/2$$, where $$\angle ACB$$ is the right angle.

The bisector $$\overline{AE}$$ splits $$\angle CAB = \alpha$$ into two.

The figure above shows the case for a relatively smaller $$r$$ (magenta) compared with $$R$$ (orange), and the figure below is for a relatively large $$r$$. As an analytic-geometry problem, I have been able to solve the coordinates $$(x_E, y_E)$$ of point $$E$$ (taking e.g. point $$A$$ as the origin), but I cannot tell from this expression if there's a sensible compass-and-straightedge construction:

$$x_E = \frac{r}{4R} \left( r + \sqrt{ r^2 + 8R^2} \right) \qquad \text{or equivalently} \qquad \cos \frac{\alpha}2 = \frac{1}{4R} \left( r + \sqrt{ r^2 + 8R^2} \right)$$

As of now I'm trying to understand the content of some related post here on StackExchange; upon first glance it seems considerable work is needed to make their results helpful for my case.

btw, this construction is related to the L'H$$\hat{o}$$pital's weight (pulley) problem (in calculus).

Here's a construction, where I'll use $s$ for the given hypotenuse: • Construct right $\triangle OAB$ with $|\overline{OA}| = |\overline{OB}| = s$.

• Construct $\overline{AC}$ perpendicular to $\overline{AB}$, with $|\overline{AC}| = r/2$.

• Construct $D$ where $\overleftrightarrow{BD}$ meets the "far side" the the circle about $C$ through $A$.

• Construct $E$, the midpoint of $\overline{BD}$.

• Construct $F$, where the circle about $B$ through $E$ meets a semicircle on $\overline{OB}$.

• Construct $G$, the reflection of $O$ across $\overline{BF}$ (which is easy, as $\angle OFB$ is necessarily a right angle).

• Construct $H$ where $\overline{BG}$ meets the semicircle on $\overline{OB}$.

• $\triangle OBH$ is the desired triangle.

Certainly, $\overline{BF}$ bisects angle $B$. That $|\overline{BX}| = r$ is trickier to establish: by the Power of a Point theorem, we have $$|\overline{OX}||\overline{XH}| = |\overline{BX}||\overline{XF}|$$ A bit of messy algebra shows that $$|\overline{BF}| = \frac{1}{4}\left(\; r + \sqrt{r^2 + 8s^2} \;\right) = \frac{1}{2}\left(\;\frac{r}{2} + \sqrt{\left(\;\frac{r}{2}\;\right)^2 + \left(\;s\sqrt{2}\;\right)^2\;}\;\right)$$ where the right-most expression gives the form that guided the construction. Then, the bisector has the correct length by virtue of the fact that the $\triangle OBH$ is, in fact, the solution.

I suspect there's a construction that makes all the relations clear.

• Amazing! I had a feeling that `power of a point' might show up at some point. Thanks. There are some stuff I'm still trying to sort out, like possibly viewing this as a construct for given length from-apex-to-orthocenter $\overline{BX}=r$ in the isosceles $\triangle OBG$ of congruent sides $\overline{OB} = \overline{GB} = s$ (instead of my original proposal of given bisector in a right triangle), meanwhile I think your solution is pretty much what I'm looking for. – Lee David Chung Lin Aug 24 '16 at 5:25
• May I ask what tool you used to make this diagram? Some features of the figure look familiar but I cannot really tell. – Lee David Chung Lin Aug 24 '16 at 5:35
• @LeeDavidChungLin: I'm glad I could help. I use GeoGebra for my figures. – Blue Aug 24 '16 at 5:53

First, construct a right triangle with legs $r$ and $R\sqrt8$, as its hypotenuse is equal to the square root in the expression: $\sqrt{r^2+8R^2}$. Constructing $\sqrt8$ and multiplying it by $R$ are reasonably short tasks, as are the remaining two in the construction of $x_E$: adding $r$ to the hypotenuse, multiplying that by $r$ and dividing the result by $4R$. Multiplication and division can be accomplished by similar triangles, while $\sqrt8$ is the diagonal of a square of side length 2. The construction of the desired right triangle follows.

• ah, yes, thanks, this procedure is manageable indeed. For this approach I wonder $\ldots$ how to arrange this construction relative to the final triangle to see geometrically how that length $x_E$ (as the horizontal component on $\overline{AB}$) makes $\overline{AE}$ the bisector? – Lee David Chung Lin Aug 22 '16 at 12:29
• None that look particularly nice. It's the same thing with the construction of the heptadecagon; proving that the polygon constructed is regular requires a lot of algebra. – Parcly Taxel Aug 22 '16 at 12:39
• ha, good point. – Lee David Chung Lin Aug 22 '16 at 12:41

Take a straight horizontal line $l$, draw a point $H$ on it and draw a segment of length $4R$ (basically intersect the line $l$ with a circle of radius $4R$ and center $H$) to the right and a segment of length $2R$ to the left. This way you get a segment $PQ \subset l$ of length $6R$ where $P$ is the left point on $l$ at distance $2R$ from $H$ and $Q$ is the right point on $l$ at distance $4R$ from $H$. Take the midpoint $M$ of $PQ$ and draw a circle $c_1$ of radius $3R$ centered at $M$ (i.e. a circle with diameter $PQ$). Now, draw the perpendicular to $PQ$ from point $H$ and let it intersect the circle $c_1$ at the point $T$ (there are two of them, just pick one, let's say the one above $PQ$). Now $HT^2 = HP\cdot HQ = 8R^2$. Next, draw on $PQ$ a point $E$ at a distance $r$ from $H$. Now, the segment $TE = \sqrt{HE^2+HT^2} = \sqrt{r^2 + 8R^2}$. Extend the segment $TE$ past $E$ and draw a point $A$ on $TE$ so that $AE = r$. Observe that $AT = AE + TE = r + \sqrt{r^2 + 8R^2}$. Now draw a circle $c_A$ centered at $A$ and of radius $4R$. Draw a line $l_T$ through point $T$ and perpendicular to line $TE$ (which is the same as line $TA$). Take one of the two intersection points of $l_T$ with $c_A$, call it $D$. By construction $$\cos{(\angle DAT)} = \frac{AT}{AD} =\frac{r + \sqrt{r^2 + 8R^2}}{4R}.$$ Draw point $B$ on the segment $AD$ so that $AB = R$. Draw the line $BE$ and construct the orthogonal projection $C$ of point $A$ on the line $BE$. Triangle $ABC$ is your triangle.

One way to construct point $C$ and to verify the construction is the correct one, is to look again at the intersection of circle $c_A$ of radius $4R$ and the line $l_T$ passing through point $T$ and perpendicular to line $AT$. One intersection point we already selected and called $D$ and let the other be $D'$. Then triangle $ADD'$ is isosceles with $AD=AD'$. Draw the altitude $DC'$ where $C'$ is on $AD'$. Define by $E'$ the intersection point of $AT$ and $DC'$ (i.e. it is the orthocenter of $ADD'$). The point $C'$ is also the second intersection point of $AD'$ and the circle circumscribed around right-angled triangle $ACT$ which is also the circle of diameter $AD$.

Now, we want to make sure that $AE' = 4r$ (so that the 4 times smaller segment $AE$ is of length $r$). By construction $AE'$ is an angle bisector of angle $\angle DAC'$. Now let for simplicity denote $$AT = h = r + \sqrt{r^2 + 8R^2} \,\, \text{ and by } \,\, AD = AD' = a = 4R.$$ Observe that $\angle DAT = \angle D'AT$ since $AT$ is the angle bisector of $\angle DAD'$. Furthermore $\angle E'TC' = \angle E'DT$ because $C'ADT$ is inscribed in a circle, or also because the right-angled triangles $TDE'$ and $C'AE'$ are similar. Thus $\angle TAD = \angle TDE'$ which means triangles $TAD$ and $TDE'$ are similar so $$\frac{AD}{AT} = \frac{DE'}{DT}.$$ Follow the calculations \begin{align} DT &= \sqrt{a^2 - h^2} \,\, \text{ and } \,\, DE' = \sqrt{(h-AE')^2 + a^2-h^2}\\ \frac{AD}{AT}&=\frac{DE'}{DT} = \frac{a}{h} =\frac{\sqrt{(h-AE')^2 + a^2-h^2}}{\sqrt{a^2 - h^2}}. \end{align}
Square the last equation on both sides \begin{align} &\frac{a^2}{h^2} = \frac{(h-AE')^2 + a^2-h^2}{a^2 - h^2} = \frac{(h-AE')^2}{{a^2 - h^2}} + 1,\\ &(h-AE')^2 = \left(\frac{a^2}{{h^2}} - 1\right)(a^2-h^2) = \frac{(a^2-h^2)^2}{h^2}\\ &AE' = h - \frac{a^2-h^2}{h} = \frac{2h^2 - a^2 }{h} \end{align}
Plug the original expressions for $a$ and $h$ in the latter equation and you get \begin{align} AE' &= \frac{2 (r^2 + r^2 + 8R^2 + 2r\sqrt{r^2+8R^2}) - 16R^2}{r+\sqrt{r^2+8R^2}}\\ &= \frac{4r^2 + 4r\sqrt{r^2+8R^2}}{r+\sqrt{r^2+8R^2}}\\ &=4r. \end{align}

• Thank you. I'm thinking that this arrangement suggests the construction of an isosceles (with base on $l_T$ and half of it being $\triangle DAT$) that naturally has the vertex angle bisected, instead of my previous line of thought involving segment of length $x_E$. How this meets the requirement of lengths $R$ and $r$ is still somewhat mysterious to me, but yeah this is helpful. – Lee David Chung Lin Aug 22 '16 at 18:11
• That is clearly true. If $D'$ is the second point of intersection of circle $c_A$ and $l_T$ (the other one being $D$) then $ADD'$ is isosceles triangle with angle the angle of the right angled triangle you are after. Then the altitude from $D$ to $AD'$ intersects $AD'$ at a point $C'$ so that the triangle $ADC'$ is a right angled triangle which is similar to the original triangle $ABC$, it's just 4 times bigger. Is this what you are trying to prove? – Futurologist Aug 22 '16 at 20:04
• Yeah, sort of. Before you mentioned it I didn't think of this approach of having $\triangle ADC'$ that is similar but 4 times the size. I guess this can be another example of the common method in plane geometry: construct an auxiliary similar shape to illuminate some facts. – Lee David Chung Lin Aug 24 '16 at 5:09