Slopefessor Fr1day – Page 18 – High School Geometry

Twenty one equivalent statements

In a bid to mirror the number of statements in the Invertible Matrix Theorem, to mimic right triangles, as well as to have some memory of $2021$

, we’ve assembled $21$

statements that are equivalent in certain triangles. Most of these statements are trivial, but what is vital is that right triangles and the triangles that satisfy the statements are rivals. Vital rivals.

Problem statement

Let $\triangle ABC$

be a non-right triangle with side-lengths $a,b,c$

, altitudes $h_A,h_B,h_C$

, circumradius $R$

, circumcenter $O$

, orthocenter $H$

, and nine-point center $N$

. Then the following statements are equivalent:

$AH=b$
$BH=a$
$CH=2h_C$
$\cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}$
$\cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}$
$\cos C=\frac{2ab}{a^2+b^2}$
$\cos^2 A+\cos^2 B=1$
$\sin^2 A+\sin^2 B=1$
$a\cos A+b\cos B=0$
$h_A^2+h_B^2=AB^2$
$\frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}$
$a^2+b^2=4R^2$
$A-B=\pm 90^{\circ}$
$(a^2-b^2)^2=(ac)^2+(cb)^2$
the orthic triangle is obtuse isosceles
radius $OC$ is parallel to side $AB$
the nine-point center lies on $AB$
the geometric mean theorem holds
the bisector of $\angle C$ has length $l$ , where $l^2=\frac{2a^2b^2}{a^2+b^2}$
segment $CH$ is tangent to the circumcircle at point $C$
the orthocenter is a reflection of vertex $C$ over side $AB$ .

The connection between the orthic triangle and the geometric mean theorem seems cool. The restriction to non-right triangles is crucal: some of the statements hold in right triangles, but not the entire chain of statements.

Our favorites: $14\&15\&17\&18\&20\&21$ . Notice the last two.

Partial solutions

A complete proof will require about $21$ implications, and we’ve already seen some of these implications in previous iterations of our discussions: May 14, May 28, June 14, and June 28. Below we make some random selections.

Suppose that $A-B=\pm 90^{\circ}$

in $\triangle ABC$

. PROVE that the radius through $C$

is parallel to side $AB$

To be specific, let $B-A=90^{\circ}$ and set $A=\theta$ . Then $B=90^{\circ}+\theta$ and $C=90^{\circ}-2\theta$ . Consider the circumcircle shown below:

The angle which the major arc $AC$ subtends at the center of the circle is twice the angle it subtends at the circumference, and so:

$\textrm{reflex}~\angle AOC=2(90^{\circ}+\theta)\implies \textrm{obtuse}~\angle AOC=180^{\circ}-2\theta$

Since $\triangle AOC$ is isosceles, we have that

$\angle OAC=\angle OCA=\theta$

This shows that $OC$ is parallel to $AB$ .

Let $O$

be the circumcenter of $\triangle ABC$

. If radius $OC$

is parallel to side $AB$

, PROVE that $A-B=\pm 90^{\circ}$

Begin by drawing the circumcircle:

If $\angle BAC=\theta$ , then $\angle OCA=\theta$ , as marked above. Reason: radius $OC$ is parallel to side $AB$ by assumption. Next, $\triangle AOC$ is isosceles, and so $\angle OAC=\theta$ . In turn, this yields $\angle AOC=180^{\circ}-2\theta$ . The major arc $AC$ now subtends an angle of $180^{\circ}+2\theta$ at the center of the circle. This means that it subtends an angle of $90^{\circ}+\theta$ at the circumference. Thus, $\angle ABC=90^{\circ}+\theta$ . The difference between $\angle BAC$ and $\angle ABC$ is then $90^{\circ}$ .

A different orientation having angle $\theta$ placed at $B$ will also yield the same conclusion.

Let $O$

and $H$

be the circumcenter and orthocenter of $\triangle ABC$

. If $A-B=\pm 90^{\circ}$

, PROVE that $OC$

is perpendicular to $CH$

Let $B-A=90^{\circ}$ and set $A=\theta$ . Then $B=90^{\circ}+\theta$ and $C=90^{\circ}-2\theta$ . Draw the circumcircle, like so:

Since $\triangle ABC$ is obtuse, its orthocenter $H$ is situated outside the triangle as shown above. Join $OC$ and $CH$ . Extend side $AB$ to meet $CH$ at $F$ . Since $F$ now becomes the foot of the altitude from $C$ , we have that $BFH=90^{\circ}$ . Since $OC$ is parallel to $AB$ , it follows that $\angle OCH=90^{\circ}$ .

Radius is perpendicular to tangent at the point of contact. Implies: $HC$ is a tangent to the circumcircle at $C$ .

Let $O$

and $H$

be the circumcenter and orthocenter of $\triangle ABC$

. If segment $CH$

is a tangent to the circumcircle at $C$

, PROVE that $a^2+b^2=4R^2$

Apply the Pythagorean theorem to the right triangle $OCH$ below:

$\begin{equation*} \begin{split} OH^2&=OC^2+CH^2\\ 9R^2-a^2-b^2-c^2&=R^2+(4R^2-c^2)\\ \therefore a^2+b^2&=4R^2 \end{split} \end{equation*}$

Let $O$

and $H$

be the circumcenter and orthocenter of $\triangle ABC$

. If segment $CH$

is tangent to the circumcircle at $C$

, PROVE that $H$

is a reflection of $C$

over side $AB$

We had $a^2+b^2=4R^2$ from the preceding example. This in turn implies $BH=a$ . So $\triangle BCH$ below is isosceles with $BH=BC$ :

Extend side $AB$ to meet $CH$ at $F$ . Since $F$ now becomes the foot of the altitude from $C$ , we have that $BFH=90^{\circ}$ . Altitude $BF$ bisects the base, so $CF=FH$ . This proves that $H$ is a reflection of $C$ over side $AB$ .

In $\triangle ABC$

, suppose that $(a^2-b^2)^2=(ac)^2+(cb)^2$

. PROVE that the nine-point center $N$

lies on side $AB$

internally or externally.

We focus on the internal case and use the segment addition postulate. Note that in any triangle $ABC$ with orthocenter $H$ , circumcenter $O$ , and nine-point center $N$ we have:

$AN^2=\frac{2AH^2+2AO^2-OH^2}{4}$

We proved before that $AH=b$ when $(a^2-b^2)^2=(ac)^2+(cb)^2$ . So

$AN^2=\frac{2AH^2+2AO^2-OH^2}{4}=\frac{2b^2+2R^2-(9R^2-a^2-b^2-c^2)}{4}$

After some simplifications, we obtain

$AN^2=\frac{(a^2-3b^2)^2}{16(a^2+b^2)}\implies AN=\frac{|(a^2-3b^2)|}{4\sqrt{(a^2+b^2)}}$

Similarly:

$BN=\frac{|(b^2-3a^2)|}{4\sqrt{(a^2+b^2)}}$

Let’s examine the absolute values. There are four cases to consider.

First, we can’t have $a^2-3b^2> 0$ and $b^2-3a^2> 0$ simultaneously. Otherwise, their sum must be greater than zero as well; but their sum is $-2(a^2+b^2)< 0$ .

Next, suppose that $3b^2-a^2> 0$ and $3a^2-b^2> 0$ . Then the sum is $2(a^2+b)^2$ , and so:

$AN=\frac{3b^2-a^2}{4\sqrt{(a^2+b^2)}},~~BN=\frac{3a^2-b^2}{4\sqrt{(a^2+b^2)}}\implies AN+BN=\frac{\sqrt{a^2+b^2}}{2}$

Because $a^2+b^2=4R^2$ , this leads to $AN+BN=R$ . This is a special case. If the points $A,B,N$ aren’t co-linear, then in $\triangle ABN$ , the median through $N$ passes through the nine-point circle, and so the length of this median is a radius of the nine-point circle, namely $\frac{R}{2}$ . We now have a triangle $ABN$ in which the sum of two sides is $R$ and a median has length $\frac{R}{2}$ . This is impossible (see the exercises at the end). Indeed, the side-lengths of $\triangle ABN$ have to be of the form $\frac{R-c}{2},\frac{R+c}{2},c$ for sides $AN,BN,AB$ (or sides $BN,AN,AB$ ). If we compute the cosine of the angle at $N$ , we obtain

$\cos N=\frac{\left(\frac{R-c}{2}\right)^2+\left(\frac{R+c}{2}\right)^2-c^2}{2\times \frac{R-c}{2}\times \frac{R+c}{2}}=1\implies\angle N=0$

The third and fourth cases are the same. For example, $3b^2-a^2> 0$ and $b^2-3a^2> 0$ . Then take

$AN=\frac{3b^2-a^2}{4\sqrt{(a^2+b^2)}},~~BN=\frac{b^2-3a^2}{4\sqrt{(a^2+b^2)}}$

and obtain

$AN+BN=\frac{4(b^2-a^2)}{4\sqrt{(a^2+b^2)}}=c$

If the nine-point center $N$

lies on $AB$

internally, PROVE that $(a^2-b^2)^2=(ac)^2+(cb)^2$

Using the fact that $AN+BN=AB=c$ , we have:

$\sqrt{\frac{2AH^2+2AO^2-OH^2}{4}}+\sqrt{\frac{2BH^2+2BO^2-OH^2}{4}}=c$

After all the trouble of the previous example, we don’t want to bother you with another seemingly lengthy procedure, but note that $(a^2-b^2)^2=(ac)^2+(cb)^2$ results after simplifications.

If the length of the bisector of $\angle C$

satisfies $l^2=\frac{2a^2b^2}{a^2+b^2}$

, PROVE that $(a^2-b)^2=(ac)^2+(cb)^2$

Normally, $l^2=ab\left(1-\left(\frac{c}{a+b}\right)^2\right)$ . So

$ab\left(1-\left(\frac{c}{a+b}\right)^2\right)= \frac{2a^2b^2}{a^2+b^2}\implies (a^2-b)^2=(ac)^2+(cb)^2$

If $(a^2-b)^2=(ac)^2+(cb)^2$

, PROVE that the length $l$

of the bisector of $\angle C$

satisfies $l^2=\frac{2ab}{a^2+b^2}$

First isolate $c^2$ from $(a^2-b)^2=(ac)^2+(cb)^2$ as $c^2=\frac{(a^2-b^2)^2}{a^2+b^2}$ and then use in the standard angle bisector formula:

$l^2=ab\left(1-\left(\frac{c}{a+b}\right)^2\right)=\frac{2a^2b^2}{a^2+b^2}.$

Peculiar scenario

Below is a reward of the labour in example 6.

Find a triangle for which the nine-point center coincides with a vertex.

Take $\triangle ABC$ with $\angle A=\angle C=30^{\circ},~\angle B=120^{\circ}$ . Then $b^2=3a^2$ . The nine-point center is precisely $B$ , since (according to example 6):

$BN=\frac{|(b^2-3a^2)|}{4\sqrt{(a^2+b^2)}}=0.$

Our next post will show that no other triangle has this property.

Takeaway

Let $\triangle ABC$ be a non-right triangle with side-lengths $a,b,c$ , altitudes $h_A,h_B,h_C$ , circumradius $R$ , circumcenter $O$ , orthocenter $H$ , and nine-point center $N$ . Then the following statements are equivalent:

$AH=b$
$BH=a$
$CH=2h_C$
$R=\frac{|a^2-b^2|}{2c}$
$h_C=R\cos C$
$\cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}$
$\cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}$
$\cos C=\frac{2ab}{a^2+b^2}$
$\cos^2 A+\cos^2 B=1$
$\sin^2 A+\sin^2 B=1$
$a\cos A+b\cos B=0$
$2\cos A\cos B+\cos C=0$
$OH^2=5R^2-c^2$
$h_A^2+h_B^2=AB^2$
$\frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}$
$a^2+b^2=4R^2$
$A-B=\pm 90^{\circ}$
$(a^2-b^2)^2=(ac)^2+(cb)^2$
$AH^2+BH^2+CH^2=8R^2-c^2$
radius $OC$ is parallel to side $AB$
segment $CH$ is tangent to the circumcircle at point $C$
the nine-point center lies on $AB$
the orthic triangle is obtuse isosceles
the geometric mean theorem holds
the bisector of $\angle C$ has length $l$ , where $l^2=\frac{2a^2b^2}{a^2+b^2}$
the orthocenter is a reflection of vertex $C$ over side $AB$ .

No need to wait until 20 26 for twenty six equivalent statements.

Tasks

(Identical traits) Consider a right triangle with side-lengths , circumradius , circumcenter , nine-point center , and . Let be the circumcenter of and let be its circumradius. Similarly, let be the circumcenter of and let be its circumradius.
- Show that $N,O_a,O_b$ are co-linear.
- PROVE that $R^2=aR_b$ and $R^2=bR_a$ .
- Deduce that the ratio in which the bisector of $\angle C$ divides side $AB$ is the same as the circumradii ratio $\frac{R_a}{R_b}$ .
(Identical traits) Consider a non-right triangle with side-lengths , circumradius , circumcenter , and nine-point center . Let be the circumcenter of and let be its circumradius. Similarly, let be the circumcenter of and let be its circumradius. If :
- Show that $N,O_a,O_b$ are co-linear.
- PROVE that $R^2=aR_b$ and $R^2=bR_a$ .
- Deduce that the ratio in which the bisector of $\angle C$ divides side $AB$ is the same as the circumradii ratio $\frac{R_a}{R_b}$ .
  (There goes a non-right triangle that mimics a right triangle in many aspects.)
(Isosceles trapezium) Suppose that an obtuse satisfies any of the equivalent statements considered in this post. Let be a point on the circumcircle such that is a diameter.
- PROVE that $AD=BC$ .
- Deduce that quadrilateral $ABCD$ is an isosceles trapezium.
(Impossible triangle) In triangle , suppose that and that the length of the median from is , as per the special case encountered in the course of example 6.
- PROVE that the lengths of sides $AN$ and $BN$ must be $\frac{R-c}{2}$ and $\frac{R+c}{2}$ (or the other way).
- Using the triangle inequality, deduce that the three points $A,N,B$ are co-linear.
PROVE that the following two statements are equivalent for any triangle :
- $\angle C=90^{\circ}$
- the length $l$ of the bisector of $\angle C$ satisfies $l^2=\frac{2a^2b^2}{(a+b)^2}$ .

Geometric mean for angle bisectors

The three traditional triangle cevians are medians, altitudes, and angle bisectors. Of these three, only the altitude is usually considered in the geometric mean theorem. However, the median also has an associated, isolated geometric mean theorem (example 1). Since what is good for the goose is also good for the angle bisector, we thought it good to present a geometric mean theorem for this third triangle cevian.

Let $a,b,c$ be the side-lengths of $\triangle ABC$ , and let $l$ be the length of the bisector of $\angle C$ , as shown below:

Then $l$ is the geometric mean of $m$ and $n$ if, and only if:

(1) $\begin{equation*} \begin{split} (a+b)^2&=2c^2 \end{split} \end{equation*}$

We’ll see the simplifications that result when equation (1) restricts to right triangles

(2) $\begin{equation*} \begin{split} c^2&=a^2+b^2 \end{split} \end{equation*}$

and the simplifications that result when (1) is considered in the context of triangles satisfying

(3) $\begin{equation*} (a^2-b^2)^2=(ac)^2+(cb)^2 \end{equation*}$

Let’s set the ball rolling with what a right triangle enjoys solely.

PROVE that a median is the geometric mean of the two segments into which it divides the opposite side if, and only if, the parent triangle is a right triangle.

Suppose that $\triangle ABC$ is a right triangle with $\angle C=90^{\circ}$ . Then the median from vertex $C$ is half of the length of the hypotenuse: $m_c=\frac{1}{2}c$ and so $m_c$ is the geometric mean of $\frac{1}{2}c$ and $\frac{1}{2}c$ . On the other hand, if a triangle has the property that one of its medians is the geometric mean of the two segments on the opposite side, for example, $m_c^2=\frac{1}{2}c\times \frac{1}{2}c$ , we show that the triangle is a right triangle. Indeed:

$\begin{equation*} \begin{split} m_c^2&=\frac{1}{2}c\times \frac{1}{2}c\\ \frac{2a^2+2b^2-c^2}{4}&=\frac{1}{4}c^2\\ 2a^2+2b^2&=2c^2\\ \therefore a^2+b^2&=c^2\\ \end{split} \end{equation*}$

So the triangle is a right triangle. Since all medians are internal, this property is unique to right triangles.

Formula derivation

Our main result is example 4, for which we may need the next two examples.

In the diagram below, $CD$

is the bisector of $\angle C$

. PROVE that $m=\frac{b}{a+b}c$

and $n=\frac{a}{a+b}c$

We first have that $m+n=c$ . Also:

$\angle ADC + \angle CDB=180^{\circ}\implies \sin \angle ADC=\sin \angle CDB$

The sine rule applied to $\triangle$ s $ADC$ and $CDB$ then gives:

$\begin{equation*} \begin{split} \frac{m}{\sin\theta}&=\frac{b}{\sin \angle ADC}\\ \frac{n}{\sin\theta}&=\frac{a}{\sin \angle CDB}\\ \therefore \frac{m}{n}&=\frac{b}{a}\\ m&=\frac{b}{a}n\\ c-n&=\frac{b}{a}n\\ c&=\left(\frac{b}{a}+1\right)n\\ c&=\frac{a+b}{a}n\\ \therefore n&=\frac{a}{a+b}c\\ \therefore m&=\frac{b}{a+b}c\\ \end{split} \end{equation*}$

Basically, $m:n=b:a$ . This is very useful in finding the coordinates of the foot of an angle bisector, which in turn is useful in finding the coordinates of the incenter of a triangle.

PROVE that $l^2=ab\left(1-\left(\frac{c}{a+b}\right)^2\right)$

, where $l$

is the length of the bisector of $\angle C$

Let’s refer to the diagram in example 2 again:

and then apply Stewart’s theorem to $\triangle ABC$ to obtain

$a^2m+b^2n=c(l^2+mn)\implies l^2=\frac{a^2m+b^2n}{c}-mn$

We now use the fact that $m=\frac{b}{a+b}c$ and $n=\frac{a}{a+b}c$ from example 2.

$\begin{equation*} \begin{split} l^2&=\frac{a^2m+b^2n}{c}-mn\\ &=\frac{a^2\times \frac{b}{a+b}c+b^2\times \frac{a}{a+b}c}{c}-\frac{b}{a+b}c\times \frac{a}{a+b}c\\ &=ab-\frac{abc^2}{(a+b)^2}\\ &=ab\left(1-\frac{c^2}{(a+b)^2}\right) \end{split} \end{equation*}$

An alternative procedure is to use area arguments as well as the cosine formula and double angle identities for $\sin 2\theta$ and $\cos 2\theta$ . This way one avoids going through Stewart’s theorem and an explicit determination of $m$ and $n$ .

In the diagram below, PROVE that the length $l$

of the bisector of $\angle C$

is the geometric mean of the segments $AD$

and $DB$

if, and only if, $(a+b)^2=2c^2$

First suppose that $l$ is the geometric mean of $m$ and $n$ . Then $l^2=mn$ .

$\begin{equation*} \begin{split} \implies ab\left(1-\left(\frac{c}{a+b}\right)^2\right)&=\frac{a}{a+b}c\times \frac{b}{a+b}c\\ 1-\frac{c^2}{(a+b)^2}&=\frac{c^2}{(a+b)^2}\\ 1&=2\frac{c^2}{(a+b)^2}\\ \therefore (a+b)^2&=2c^2\\ \end{split} \end{equation*}$

Conversely, suppose that $(a+b)^2=2c^2$ .

$\begin{equation*} \begin{split} l^2&= ab\left(1-\left(\frac{c}{a+b}\right)^2\right)\\ &=ab\left(1-\frac{c^2}{2c^2}\right)\\ &=\frac{1}{2}(ab)\\ mn&=\frac{b}{a+b}c\times \frac{a}{a+b}c\\ &=\frac{abc^2}{(a+b)^2}\\ &=\frac{abc^2}{2c^2}\\ &=\frac{1}{2}ab\\ \therefore l^2&=mn \end{split} \end{equation*}$

Notice how we obtain another equivalence: $l^2=mn\iff (\sqrt{2}l)^2=ab$ . Translate into words: we killed two birds with one stone.

For a quick example consider triangle $ABC$

with vertices at $A(0,0)$

, $B\left(6\sqrt{2},0\right)$

, and $C\left(2\sqrt{2},\sqrt{17}\right)$

, shown below:

The foot $D$ of the bisector of $\angle C$ has coordinates $\left(\frac{5}{2}\sqrt{2},0\right):=D$ . By the distance formula:

$l^2=\left(\frac{5}{2}\sqrt{2}-2\sqrt{2}\right)^2+\left(0-\sqrt{17}\right)^2=\frac{35}{2}.$

Also:

$m=\frac{5}{2}\sqrt{2},~n=\frac{7}{2}\sqrt{2}\implies mn=\frac{35}{2}=l^2.$

Thus, $CD$ is the geometric mean of $AD$ and $DB$ .

Fifteen degrees

We now restrict equation (1) to right triangles and to triangles satisfying equation (3) . It turns out that the interior angles are all multiples of $15^{\circ}$ .

If the bisector of $\angle C=90^{\circ}$

in a right triangle satisfies equation (1), PROVE that the triangle is isosceles.

Using the fact that $a^2+b^2=c^2$ and $(a+b)^2=2c^2$ , we have:

$(a+b)^2=2(a^2+b^2)\implies (a-b)^2=0\implies a=b$

The angles of the triangle will then be $45^{\circ}$ , $45^{\circ}$ , $90^{\circ}$ . All multiples of $15^{\circ}$ .

If the bisector of the apex $\angle C$

in an isosceles triangle satisfies equation (1), PROVE that the triangle is a right triangle.

In this case $a=b$ . Together with equation (1), we have

$(a+b)^2=2c^2\implies (2a)^2=2c^2\implies c^2=2a^2=a^2+b^2$

Alternatively, the bisector is a median in this case, and so the conclusion follows from example 1 and example 4.

Suppose that $\triangle ABC$

satisfies equation (3). If the bisector of $\angle C$

satisfies equation (1), PROVE that $a^2-4ab+b^2=0$

We have: $(a+b)^2=2c^2\& (a^2-b^2)^2 =c^2(a^2+b^2)$ . Isolate $c^2$ and simplify:

$(a+b)^2=2\left(\frac{(a^2-b^2)^2}{a^2+b^2}\right)\implies a^2+b^2=2(a-b)^2\implies \scriptstyle a^2-4ab+b^2=0$

Suppose that $\triangle ABC$

satisfies equation (3). If the bisector of $\angle C$

satisfies equation (1), PROVE that $\angle A=15^{\circ}$

, $\angle B=105^{\circ}$

, $\angle C=60^{\circ}$

Since $\cos C=\frac{2ab}{a^2+b^2}$ (see here), we have (by example 7 above):

$\cos C=\frac{2ab}{4ab}=\frac{1}{2}\implies \angle C=60^{\circ}$

Also, $\cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}$ and $\cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}$ yield $\angle A=105^{\circ}$ and $\angle B=15^{\circ}$ (or $\angle A=15^{\circ}$ and $\angle B=105^{\circ}$ ).

Further developments

Here are two out of a few more consequences.

Let $a,b,c$

be the side-lengths of $\triangle ABC$

. If equations (1) and (3) are satisfied, PROVE that the circumradius $R$

is the geometric mean of $a$

and $b$

We have $a^2+b^2=4R^2$ and $a^2-4ab+b^2=0$ . Eliminate $a^2+b^2$ to obtain $R^2=ab$ .

If $\triangle ABC$

satisfies equations (1) and (3), PROVE that the squared lengths $h_C^2,l^2,R^2$

form a geometric progression. Here, $l$

is the length of the bisector of $\angle C$

, $h_C$

is the length of the altitude from vertex $C$

, and $R$

is the circumradius.

In example 4 we had $l^2=\frac{1}{2}(ab)$ ; in example 9 above we had $R^2=ab$ . Now let’s calculate $h_C$ :

$h_C=\frac{ab}{2R} =\frac{ab}{\sqrt{a^2+b^2}}=\frac{ab}{\sqrt{4ab}}\implies h_C^2=\frac{1}{4}(ab)$

We see that $h_C^2,l^2,R^2$ form a geometric progression with common ratio $r=2$ .

Takeaway

In triangle $ABC$ , let $D$ be the foot of the bisector of $\angle C$ . Then the two statements below are equivalent:

$CD$ is the geometric mean of $AD$ and $DB$
$\sqrt{2} CD$ is the geometric mean of $AC$ and $CB$ .

Further, for any triangle $ABC$ whose side-lengths $a,b,c$ satisfy equation (1), the following two statements are equivalent:

$\triangle ABC$ is a right triangle
$a^2-2ab+b^2=0$ .

Finally, for any triangle $ABC$ whose side-lengths $a,b,c$ satisfy equation (1), the two statements below are equivalent:

$\triangle ABC$ is a pseudo right triangle
$a^2-4ab+b^2=0$ .

Tasks

Using with vertices at , , , verify that:
- its incenter is $I\left(-3+\sqrt{5},-1+\sqrt{5}\right)$ .
- the ratio of the coordinates $(x,y)$ of the incenter is golden; that is, it satisfies $x^2-xy-y^2=0\implies \left(\frac{x}{y}\right)^2-\frac{x}{y}-1=0$ .
- the slope of the bisector of $\angle B$ is $-\frac{1+\sqrt{5}}{2}$ , the negative of the golden ratio.
Let be the side-lengths of triangle .
- PROVE that the harmonic mean of $\frac{ac}{a+b}$ and $\frac{bc}{a+b}$ is $\frac{2abc}{(a+b)^2}$ .
- Deduce that the bisector of $\angle C$ is the harmonic mean of the two segments on the opposite, if $(a+b)^2=c^2+2c$ .
Suppose that the side-lengths of triangle satisfy equation (3), namely .
- PROVE that the length $l$ of the bisector of $\angle C$ satisfies $l^2=\frac{2a^2b^2}{a^2+b^2}$ .
- Deduce that $l^2$ is the harmonic mean of $a^2$ and $b^2$ .
Suppose that satisfies the usual Pythagorean identity .
- If the bisector of $\angle A$ satisfies equation (1) adapted as $(b+c)^2=2a^2$ , PROVE that $a^2=8b^2$ .
- Under the above, deduce that $\sin A=\frac{2\sqrt{2}}{3}$ , $\sin B=\frac{1}{3}$ .
In $\triangle ABC$ , let the side-lengths be $a,b,c$ , and let $R,h_C,l$ denote the circumradius, altitude from $C$ , and the length of the bisector of $\angle C$ . If $\triangle ABC$ satisfies equation (3), PROVE that the squared lengths $h_C^2,l^2,R^2$ form a geometric progression with common ratio $r=2$ .
In $\triangle ABC$ , let $R,h_C,m_C$ denote the circumradius, altitude from $C$ , and the median from $C$ . If $\triangle ABC$ satisfies equations (1) and (3), PROVE that $m_C^2=h_C^2+R^2$ .
(Basically a right triangle can be formed with the median $m_C$ as the hypotenuse.)