Prove two angles add up to 90 degrees

Question

$\triangle ABC$ is inscribed within circle $O$. $D$ is the midpoint of $AC$. $E$ is on $AB$ such that $ED/EB=CD/CB$. $CE$ intersects circle $O$ at $F$. Prove that $\angle EDF + \angle CDB = 90^{\circ} $.

---

The condition $ED/EB=CD/CB$ is awkward. I am thinking of using Menelaus theorem on $\triangle ABC$ and line segment $DE$ because there are a lot of equal line segments and ratios, but I didn't go very far.

Answer 1

There are two possibilities to construct the point $E$. Both start by constructing the point $X\in BD$, so that $X$ is the angle bisector of $\hat C$ in $\Delta ABC$.

• Because of $\frac {XD}{XB}=\frac{CD}{CB}=\frac{DD}{EB}$ (angle bisector theorem, and the given equality of proportions) $ED$ is the angle bisector in $\Delta EDB$. So let us construct points $E$ with this property. Consider the circle $(X)$ centered in $X$ which is tangent to the line $AB$. Draw the two tangents from $D$ to this circle, they intersect $AB$ in two points, $E,E'$, both being possible choices for the point $E$ in the problem.

• A second possibility uses the fact that the geometric locus of all points $P$ with given constant value for the ratio $k=PD/PB$ is a circle. This circle has the line $DB$ as symmetry. The simplest way to see this is to make use a system of coordinates with $D,B$ in $(-1,0)$ and $(1,0)$, then rewrite the given relation as $\displaystyle k^2=\frac{(x+1)^2+y^2}{(x-1)^2+y^2}$. In our case, this circle passes through $X, X'$ the points where the inner, respectively outer angle bisector of $\hat C$ in $\Delta ABC$ intersect $BD$. This is a symmetry axis, so $XX'$ is a diameter in this circle, let $\Xi$ be its center, the mid point of $XX'$. We denote this circle by $(\Xi)$. It intersects $AB$ in two points $E,E'$. (We let $E$ be the point closer to $B$.) See also

https://en.wikipedia.org/wiki/Circle#Circle_of_Apollonius

---

The second possibility is maybe better to work with, so we restate the problem equivalently, avoiding the annoying metric condition. As i always proceed in such situations, my solution is not the quick solution going straightforward to the needed conclusion, instead, all "interesting" properties in the given geometric constellation (related or not to the target property) are listed and shown. A long time i walked well with this strategy, the understanding of the problems is optimal, alternative solutions are possible. So let us state and prove the following...

---

Problem: Let $\Delta ABC$ be a triangle. We introduce the following points.

• $O$ is its circumcenter.
• $D$ is the mid point of $AB$.
• $X,X'$ are the intersections of the inner, respectively outer angle bisector with $DB$. Let $\Xi$ be the mid point of the segment $XX'$.
• $E,E'$ are the intersections of the circle $(\Xi)$ of diameter $XX'$ with the line $AB$. (Let $E$ be closer to $B$ to fix notations.)
• $CE$, $CE'$ intersect the circumcircle $(O)$ of $\Delta ABC$ in $F,F'$ respectively.
• $T$ is $XE\cap X'E'$.
• $H$ is the orthocenter of $\Delta TXX'$.
• $L,L'$ are the intersections of $DO$ with the circumcircle $(O)$.
• Let $U$ be $LF\cap L'F'$. Let $R$ be the orthocenter in $\Delta ULL'$.

Then we have:

• (1) $THD$ is the third height in $\Delta TXX'$. Moreover, $EE'\Xi D$ cyclic. As a corollary, $TD$ bisects $\widehat{EDE'}$ and each of the halves is $\widehat{ECE'}$.
• (2) $FF'$, $THD$, and $E'AEB$ are concurrent in a point $S$.
• (3) $UD$ is the third height in $\Delta ULL'$, and $UD=UARDC$. Moreover, $FF'DO$ cyclic. As a corollary, $AD$ bisects $\widehat{FDF'}$ and each of the halves is $\widehat{FCF'}$.
• (4) $THD$ is the angle bisector of the angle $\widehat{EDE'}$.
• (5) $ARD$ is the angle bisector of the angle $\widehat{FDF'}$.
• (6) The angles $\widehat{EDF}$, $\widehat{E'DF'}$, $\widehat{TDA}$, and the angle between $BXD\Xi X'$ and the diameter $LODL'$ are equal. (The last angle is the complement of $\widehat{CDB}$, thus solving the issue in the OP.)

---

A picture:

---

Proofs:

---

(1) Because $XX'$ is a diameter in $(\Xi)$ we have $XE'\perp X'E'$, $XE\perp X'E$, so $XE'$, $X'E$ are heights in $\Delta TXX'$. Which is the third height? $EX$ bisects $\widehat{DEB}$ and $EXX'E'$ cyclic. This implies: $$ \widehat{DEX}= \widehat{XEB}= \widehat{BX'E'}= \widehat{XX'T} \ , $$ so $EDX'X$ cyclic, so $\widehat{TDX'}=\widehat{TEX'}=90^\circ$. This was (1).

But in the same place we insert a picture of the Euler circle in the triangle $\Delta TXX'$, a circle passing through the feet of the heights, $E, E', D$, but also throught the mid point $\Xi$ of the base $XX'$.

---

(4) This is the usual property of the height, $$ \widehat{TDE}= \widehat{HDE}= \widehat{HXE}= \widehat{E'XT}= 90^\circ-\hat{T}= \widehat{EX'T}= \widehat{HX'E'}= \widehat{HDE'}= \widehat{TDE'} $$

Alternatively consider the mid point of $TH$, and use the fact that $\Xi$ and this point determine a diameter in the Euler circle, which is perpendicular on $EE'$. As an observation, we may write using the Euler circle and the circle $(\Xi)$: $$ \widehat{EDE'}= \widehat{E\Xi E'}= 2\widehat{EXE'}= 2\widehat{EX'E'}= 2\widehat{ECE'}= \ . $$

---

(3) It is more or less "the same situation" as in (1), but the points $F,F'$ are more complicated. Since $D$ is the midpoint of $AC$, we have $DO\perp ADC$. The diameter of $(O)$ on the line $DO$ is $LL'$. So $\Delta LL'F$ and $\Delta LL'F'$ have a right angle opposed to $LL'$. Then $LF'$, $L'F$ are two heights in $\Delta ULL'$. Which is the third height? We want to show it is $UD$. To have an immediate parallel to (1) here is the triangle $\Delta ULL'$ and its Euler circle, showing what we want to show:

An analytic solution is given, since i also need the ingredients for (2).

We use barycentric coordinates in the triangle $\Delta ABC$. Its side lengts are $a,b,c$ say, by usual notation. The computations will use standard notations, please consult

Barycentric coordinates for the impatient, Max Schindler and Evan Chen

for the details.

Some coordinates and equations are immediate. $$ \begin{aligned} A &= [1:0:0]=(1,0,0)\ ,\\ B &= [0:1:0]=(0,1,0)\ ,\\ C &= [0:0:1]=(0,0,1)\ ,\\ D &= \frac 12(A+C)=[1:0:1]=\frac 12(1,0,1)\ ,\\ X &= \frac {BC}{BC+CD}D+\frac {CD}{BC+CD}B =\frac{2a}{2a+b}\cdot\frac 12(1,0,1)+\frac{b}{2a+b}\cdot(0,1,0)\\ &=[a:b:a]=\frac 1{2a+b}(a,b,a)\ ,\\ &\text{... or use formally $(D-X)=-\frac b{2a}(B-X)$, i.e. $\frac{XD}{XB}=-\frac b{2a}$.}\\ &\text{... and for $X'$ formally $(D-X)=+\frac b{2a}(B-X)$, i.e. $\frac{XD}{XB}=+\frac b{2a}$,}\\ &\text{... so $X'$ is formally obtained from $X$ via $b\to-b$,}\\ X' &= \frac {BC}{BC+CD}D+\frac {CD}{BC+CD}B =\frac{2a}{2a+b}\cdot\frac 12(1,0,1)+\frac{b}{2a+b}\cdot(0,1,0)\\ &=[a:b:a]=\frac 1{2a+b}(a,b,a)\ ,\\ (\Xi) &=\text{the circle through $C,X,X'$, it has the equation}\\ 0 &= -a^2yz -b^2zx -c^2xy+(ux+vy)(x+y+z)\ ,\qquad\text{ where}\\ u &=\frac 1{4a^2-b^2}\cdot b^2(a^2-c^2)\ ,\\ v &=\frac 1{4a^2-b^2}\cdot a^2(2a^2-b^2+2c^2)\ ,\\ E,E'&\text{ are the two points $E_\pm$ in the intersection of $(\Xi)$ with $AB$ $(z=0)$,}\\ E_+ &= [1:m_+^E:0]=(x_+^E, y_+^E, 0)=(x_+^E, m_+^Ex_+^E, 0) =\frac 1{1+m_+^E}(1,m_+^E,0) \ ,\\ E_- &= [1:m_-^E:0]=(x_-^E, y_-^E, 0) =(x_-^E, m_-^Ex_-^E, 0) =\frac 1{1+m_-^E}(1,m_-^E,0) \ ,\\ &\qquad\text{where $m_\pm$ are the two roots of the equation in $M$}\\ &\qquad\text{obtained by setting $z=0$ and $y=Mx$ in $(\Xi)$:}\\ 0 &= -c^2 m+(u+mv)(1+m) \\ &=vm^2-(c^2-u-v)m+u\ ,\text{ so}\\ \Sigma &=m_+ + m_- = \frac 1v{(c^2-u-v)}=-\frac{2(a^2-c^2)}{2a^2-b^2+2c^2}\ ,\\ \Pi &=m_+ m_- = \frac uv=\frac{b^2(a^2-c^2)}{a^2(2a^2-b^2+2c^2)} =\frac{b^2}{2a^2}\Sigma\ ,\\ L,L' &= L_\pm =\text{ intersection of $(O)$ with perpendicular in $D$ on $AB$,}\\ L_+ &=[a(a+c) : -b^2: c(a+c)]\ ,\\ L_- &=[a(a-c) : -b^2: c(c-a)]\ ,\\ &=\qquad\text{ (and observe that $+c\leftrightarrow -c$ exchanges formally $L_+\leftrightarrow L_-$.)}\\ S &=\text{ intersection of $AB$ ($z=0$) with perpendicular in $D$ on $DB$}\\ &=\text{ solution of $x+y+z=1$, $z=0$, and}\\ &=\text{ (EFFT) for displacements $\frac 12(2x-1,2y,2z-1)$ and $\frac 12(1, -2, 1)$, so}\\ S &= [2a^2-b^2+2c^2:c^2-a^2:0]\ . \end{aligned} $$ The discriminant of the equation in $M$ is not a square in the fraction field of the ring $\Bbb[a,b,c]$, so we try not to write $m$ explicitly. Now we compute $F_\pm$, which is the solution of the equations in $(x,y,z)$: $$ \left\{ \begin{aligned} 0 &=a^2yz+b^2zx+c^2xy\ ,\\ m_\pm =\frac{y^E_\pm}{x^E_\pm} & =\frac yx\ ,\\ 1 &=x+y+z\ . \end{aligned} \right. $$ The solution of this system, where the second equation is $y=mx$ is $$ [ma^2+b^2\ :\ m(ma^2+b^2)\ :\ -mc^2]\ . $$ We obtain $F_\pm$ by setting $m_\pm$ instead of $m$. The equations for the lines $LF_+$ and $L'F_-$ are: $$ LF_+\ :\ \begin{vmatrix} a(a-c) & -b^2 & c(c-a)\\ m_+a^2+b^2 & m_+(m_+a^2+b^2) & -m_+c^2\\ x & y & z \end{vmatrix} =0 \ ,\qquad L'F_-\ :\ \begin{vmatrix} a(a+c) & -b^2 & c(c+a)\\ m_-a^2+b^2 & m_-(m_-a^2+b^2) & -m_-c^2\\ x & y & z \end{vmatrix} =0 $$ And we want to show they intersect in a point $U\in AC$, so $y_U=0$. We add the equation $x+y+z=1$ to the above two, determining $U$, and the second component is by Kramer's rule: $$ \begin{vmatrix} \color{blue}{ \begin{vmatrix} -b^2 & c(c-a)\\ m_+(m_+a^2+b^2) & -m_+c^2\\ \end{vmatrix}} & 0 & \color{magenta}{ \begin{vmatrix} a(a-c) & -b^2 \\ m_+a^2+b^2 & m_+(m_+a^2+b^2) \\ \end{vmatrix}} \\ \color{magenta}{ \begin{vmatrix} -b^2 & c(c+a)\\ m_-(m_-a^2+b^2) & -m_-c^2\\ \end{vmatrix}} & 0 & \color{blue}{ \begin{vmatrix} a(a+c) & -b^2 \\ m_-a^2+b^2 & m_-(m_-a^2+b^2) \\ \end{vmatrix}} \\ 1 & 1 & 1 \end{vmatrix} $$ (divided by the "determinant of the system"). So it is enough to show that the product
$ \color{blue}{P_\searrow}$ of the blue terms is the product $ \color{magenta}{P_\nearrow}$ of the purple terms. (So the products are invariant w.r.t. the "Galois substitution" $m_+\leftrightarrow m_-$ done in the same time with $c\leftrightarrow -c$.) $$ \begin{aligned} \color{blue}{P_\searrow} &= \color{blue}{ \begin{vmatrix} -b^2 & c(c-a)\\ m_+(m_+a^2+b^2) & -m_+c^2\\ \end{vmatrix} \cdot \begin{vmatrix} a(a+c) & -b^2 \\ m_-a^2+b^2 & m_-(m_-a^2+b^2) \\ \end{vmatrix}} \\ &= m_+\cdot \begin{vmatrix} -b^2 & c(c-a)\\ m_+a^2+b^2 & -c^2\\ \end{vmatrix} \cdot (m_-a^2+b^2) \cdot \begin{vmatrix} a(a+c) & -b^2 \\ 1 & m_- \\ \end{vmatrix} \\ &= m_+\cdot \begin{vmatrix} -b^2 & c(c-a)\\ m_+a^2 & -ac\\ \end{vmatrix} \cdot (m_-a^2+b^2) \cdot \begin{vmatrix} a(a+c) & -b^2 \\ 1 & m_- \\ \end{vmatrix} \\ &= m_+\cdot ac\begin{vmatrix} a(a-c) & -b^2\\ 1 & m_+\\ \end{vmatrix} \cdot (m_-a^2+b^2) \cdot \begin{vmatrix} a(a+c) & -b^2 \\ 1 & m_- \\ \end{vmatrix} \\ &= ac\;\color{red}{m_+\;(m_-a^2+b^2)} \;(a(a-c)m_+ + b^2)\;(a(a+c)m_- + b^2)\ ,\\[3mm] &\qquad\text{ and similarly}\\[3mm] \color{magenta}{P_\nearrow} &= \color{magenta}{ \begin{vmatrix} -b^2 & c(c+a)\\ m_-(m_-a^2+b^2) & -m_-c^2\\ \end{vmatrix} \cdot \begin{vmatrix} a(a-c) & -b^2 \\ m_+a^2+b^2 & m_+(m_+a^2+b^2) \\ \end{vmatrix}} \\ &= m_- \begin{vmatrix} -b^2 & c(c+a)\\ m_-a^2+b^2 & -c^2\\ \end{vmatrix} \cdot (m_+a^2+b^2) \begin{vmatrix} a(a-c) & -b^2 \\ 1 & m_+ \\ \end{vmatrix} \\ &= m_- \begin{vmatrix} -b^2 & c(c+a)\\ m_-a^2 & ac\\ \end{vmatrix} \cdot (m_+a^2+b^2) \begin{vmatrix} a(a-c) & -b^2 \\ 1 & m_+ \\ \end{vmatrix} \\ &= m_-\cdot a(-c) \begin{vmatrix} a(c+a) &-b^2 \\ 1 & m_- \\ \end{vmatrix} \cdot (m_+a^2+b^2) \begin{vmatrix} a(a-c) & -b^2 \\ 1 & m_+ \\ \end{vmatrix} \\ &=ac\; \color{red}{(-m_-)\;(m_+a^2+b^2)}\;(a(a+c)m_- + b^2)\;(a(a-c)m_+ + b^2) \ . \end{aligned} $$ We have the equaltiy of the terms marked in red, because of $$ m_+\;(m_-a^2+b^2) + m_-\;(m_+a^2+b^2) =2a^2\underbrace{m_+m_-}_\Pi +b^2\underbrace{(m_+ + m_-)}_\Sigma=0\ , $$ since $\Pi$ is $-\frac{b^2}{2a^2}\Sigma$. The other two factors are matching each other mot-a-mot.

$\square$

To recapitulate, we showed that $UD$ is the height of $\Delta ULL'$ (and the other two heights are $FL'$ and $F'L$), by computing the intersection $LF_+\cap L'F_-$, and showing that its $y$-component vanishes. (We can implant then either $F,F'$ or $F',F$ for the values $F_+,F_-$.) This shows (3) analytically.

---

(5) This is parallel to (4), it is the usual property of the height, $$ \widehat{ADF}= \widehat{RDF}= \widehat{RLF}= \widehat{F'LF}= 90^\circ-\hat{U}= \widehat{F'L'F}= \widehat{F'L'R}= \widehat{F'DR}= \widehat{F'DA} $$ or we can use the Euler circle in the same way as above, considering its diameter from $O$ to the mid point of $UR$, here $R$ is the orthocenter in $\Delta ULL'$. As an observation, we may write using the Euler circle and the circumcircle $(O)$: $$ \widehat{FDF'}= \widehat{FOF'}= 2\widehat{FLF'}= 2\widehat{FL'F'}= 2\widehat{FCF'}\ . $$

---

(6) This is what the OP is asking for. So far we have used intentionally two different marks for the pairs of equal (unsigned) angles in $D$: $$ \widehat{EDT}= \widehat{E'DT} \text{ and } \widehat{FDA}= \widehat{F'DA}\ . $$ They are equal, because using the circles $(O)$ and $(\Xi)$ they "united in $C$, being equal to $$ \gamma:= \widehat{ECE'}= \widehat{F'CF'}\ . $$

A rotation around $D$ with angle $\gamma$ brings the rays $DE$ into $DT$, and $DF$ into $DA$, so $\widehat{EDF}=\widehat{TDA}$. A further $\gamma$-rotation around $D$ shows
$\widehat{EDF}=\widehat{TDA}=\widehat{E'DF'}$. Using $DB\perp DT$ and $DL\perp DA$, a $90^\circ$ rotation around $D$ gives $\widehat{BDL}=\widehat{TDA}$.

This shows the OP: $$ 90^\circ-\widehat{CDB}= \widehat{BDL}= \widehat{EDF}= \widehat{E'DF'}\ . $$

---

It remains (2), a bonus in the given constellation. We use again barycentric coordinates. Recall that the points $F_\pm$ are $$ [m_\pm a^2+b^2\ :\ m_\pm(m_\pm a^2+b^2)\ :\ -m_\pm c^2]\ . $$ Then the line $F_+F_-$ has the equation $$ \begin{vmatrix} m_+ a^2+b^2\ &\ m_+(m_+a^2+b^2)\ &\ -m_+c^2\\ m_- a^2+b^2\ &\ m_-(m_-a^2+b^2)\ &\ -m_-c^2\\ x & y & z \end{vmatrix} =0\ . $$ Let $S=(x_S,y_S,z_S)$ be the intersection $F_+F_-\cap AB$. From $S\in AB$ we have $z_S=0$, so $x_S+y_S=1$, and expanding the above determinant w.r.t. the third line: $$ \begin{vmatrix} m_+(m_+a^2+b^2)\ &\ m_+\\ m_-(m_-a^2+b^2)\ &\ m_- \end{vmatrix} x_S - \begin{vmatrix} m_+ a^2+b^2\ &\ m_+\\ m_- a^2+b^2\ &\ m_- \end{vmatrix} y_S =0\ . $$ (The last column was simplified, it appears without the factor $-c^2$.) In the coefficient of $x_S$ we linearly extract on the lines the factors $m_\pm$. Then the new second column $1,1$ is used to linearly get rid of $+b^2,+b^2$ from the first column. Then in the new first column we extract the factor $a^2$.

In the coefficient of $y_S$ we use the second column $m_+,m_-$ to linearly get rid of $m_+a^2,m_-a^2$ from the first column. Then in the new first column we extract the factor $a^2$. This gives: $$ m_+m_-a^2 \begin{vmatrix} m_+\ &\ 1\\ m_-\ &\ 1 \end{vmatrix} x_S - b^2 \begin{vmatrix} 1\ &\ m_+\\ 1\ &\ m_- \end{vmatrix} y_S =0\ . $$ Recall that $m_+m_-=\Pi=\frac uv=\frac{b^2(a^2-c^2)}{a^2(2a^2-b^2+2c^2)} $, so after simplifying the above to $m_+m_-a^2x_S+b^2y_S=0$ and plugging in the formula for $m_+m_-=\Pi$ we obtain: $$ (a^2-c^2)x_S+(2a^2-b^2+2c^2)y_S=0\ , \ x_S+y_S=0\ , $$ and the solution is $$ y_S=-\frac{a^2-c^2}{a^2-b^2+3c^2}\ ,\ x_S=1-y_S\ , $$ so we compute the displacement vectors $$ \begin{aligned} DS &= S-D = (x_S,y_S,0)-\frac 12(1,0,1) =\left(x_S-\frac 12,y_S,-\frac 12\right) \\ &=\left(\frac 12-y_S,y_S,-\frac 12\right) \sim[1-2y_S\ :\ 2y_S\ :\ -1]\ ,\\ DB &= B-D =(0,1,0)-\frac 12(1,0,1) \\ &=-\frac 12(1,-2,1) \sim[1\ :\ -2\ :\ 1]\ . \end{aligned} $$ The perpendicularity $DS\perp DB$ is then equivalent to (EFFT): $$ a^2(2y_S\cdot 1+(-1)\cdot(-2) ) + b^2((1-2y_S)\cdot 1+(-1)\cdot 1 ) + c^2((1-2y_S)\cdot (-2)+2y_S\cdot1 ) =0\ . $$ which is $$ y_S(2a^2-2b^2+6c^2)+(2a^2-2c^2)=0\ . $$ Which is true.

$\square$

---

A final remark: All points (1) to (6) are now proven. Using barycentric coordinates was leading to "straightforward" solutions for (2), (3). (Using a computer algebra system, "straightforward" becomes straightforward. Here, there is some effort to type the solution, on paper things are simpler.)

If analytic = computational solutions "must be" avoided, then one needs a proof for (3), or at least for a simple part in it, for instance it is enough to show $FF'DO$ cyclic. (Note that $O$ is on the side bisector of $FF'$.) For the bonus point (2) projective geometry is maybe the path, but i could not find a proof (using Desargues, Pappus, Pascal, et caetera) in time.

But also note that barycentric coordinates are a strong tool in competitions (Olympiads, although then one should also prove formulas like EFFT for a full point harvest). Here is then an explicit example how this works in practice. No details were omitted.

I will still search for synthetic solutions, but now i have to submit, time.

Answer 2

Comment:

In figure $BG||AC$ and $\widehat {DIB}=90^o$. So $(\widehat{BDI=IDG}) +\widehat {DBI}=90^o$. Now we have to show $\widehat {BDI}=\widehat {EDF}$.As can be seen in figure $\angle HDI=\angle FDG$. But this has to be proved by angle chasing. Or we must use relation $DE/EB=DC/BC$.

Answer 3

This is the way I've used the sides-ratio condition but it didn't went too far.

Rotate point $D$ about $C$ and $E$ to $D_1$ and $D_2$ respectively such that $B-C-D_1$ and $B-E-D_2$. If follows that $\displaystyle\frac{CB}{CD_1}=\frac{EB}{ED_2}$.

In $\triangle D_1D_2B$, by converse BPT, we have $D_1D_2||CE$.