Let $ABCD$ be a convex quadrilateral. Let $|\measuredangle ABC|$=$|\measuredangle ACD|$ and $|\measuredangle ACB|$=$|\measuredangle ADC|$. Let $O$ be the circumcenter of triangle $BCD$, distinct from $A$. Prove, that $|\measuredangle OAC|$ is a right angle.

I'm supposed to use spiral similarity with center A and midpoints of sides $BC$ and $CD$ (let them be $M_A$ and $M_B$, respectively) to show that $AOM_ACM_B$ is a cyclic pentagon, but i don't know how to proceed with it. How do I do this?

  • $\begingroup$ There is something wrong with your set up... Suppose $ABCD$ is a rhombus with $60-120 - 60- 120$ angles. Place $A, C$ at the $120$ angles. Then $\angle ABC = 60 = \angle ACD$ and $\angle ACB = 60 = \angle ADC$ then $O$ lies on $AC$ and $\angle OAC$ is $0$ $\endgroup$ – Doug M Nov 29 '17 at 18:20
  • $\begingroup$ Right, there was a condition $O \neq A$, sorry about that. $\endgroup$ – Tomáš Hulla Nov 29 '17 at 18:40
  • $\begingroup$ Can someone help me please? $\endgroup$ – Tomáš Hulla Nov 30 '17 at 5:01
  • $\begingroup$ How are angles supposed to have absolute values? $\endgroup$ – user818241 Sep 21 '20 at 7:46

Particularly in the absence of better ideas, I like coordinates. Without loss of generality (as the setup is invariant under similarity transformations), you can use $A$ as the origin and $\overrightarrow{AB}$ as first unit vector. So you get

$$A=\begin{pmatrix}0\\0\end{pmatrix}\qquad B=\begin{pmatrix}1\\0\end{pmatrix}\qquad C=\begin{pmatrix}x\\y\end{pmatrix}$$

for some $x,y\in\mathbb R$. This defines the spiral similarity transformations, and applying that similarity to $C$ you get

$$D=\begin{pmatrix}x&-y\\y&x\end{pmatrix}\cdot \begin{pmatrix}x\\y\end{pmatrix}= \begin{pmatrix}x^2-y^2\\2xy\end{pmatrix}\;.$$

The midpoints are thus

$$M_A=\tfrac12(B+C)=\frac12\begin{pmatrix}x+1\\y\end{pmatrix}\qquad M_B=\tfrac12(C+D)=\frac12\begin{pmatrix}x+x^2-y^2\\y+2xy\end{pmatrix}$$

and intersecting the orthogonal lines in these points you find that the circumcenter is


To show that $\overrightarrow{AC}$ and $\overrightarrow{AO}$ are orthogonal, you check that the scalar product between these two is zero:

$$\langle C,O\rangle=\frac{x^2+y^2-1}{2y}\left\langle \begin{pmatrix}x\\y\end{pmatrix}, \begin{pmatrix}-y\\x\end{pmatrix} \right\rangle=0\;.$$


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