# Proof that 4 points lie on a circle and that center of this circle lies on the circumcircle of $\triangle ABC$

Given is acute triangle $$ABC$$. Let $$D$$ be foot of altitude from vertex $$A$$. Let $$D_1$$ be a point so that line of symmetry between $$D_1$$ and $$D$$ is line $$AB$$. Let $$D_2$$ be a point so that line of symmetry between $$D_2$$ and $$D$$ is line $$AC$$. Let points $$E_1, E_2$$ be on line $$BC$$ so that $$D_1E_1 \parallel AB$$ and $$D_2E_2 \parallel AC$$. Proof that points $$D_1, E_1, D_2, E_2$$ lie on same circle and that center of this circle lies on the circumscribed circle of triangle $$ABC$$.

My plan was to first prove that $$D_1E_1E_2D_2$$ is cyclic quadrilateral, in other words that $$\angle E_1E_2D_2 + \angle E_1D_1D_2 = 180°$$ or that $$\angle E_1D_2D_1 = \angle E_1E_2D_1$$. This would mean that points $$D_1, E_1, D_2, E_2$$ lie on same circle. However, without success.
Can someone help me with this please?

• Also , B and C are circumcenters. – Takahiro Waki Dec 3 '18 at 23:18

## 2 Answers

put $$\angle BAD=\angle EAC, \angle CAD=\angle ECB$$.

EB is perpendicular $$D_1E_1$$. Since $$BD_1E_1$$ is isosceles triangle(since those angles are equal), $$ED_1E_1$$ is isosceles triangle, too. E is on two perpendiculars of $$D_1E_1$$ and $$D_2E_2$$. This shows that E is the circumcenter of these four points.

• Excellent answer, though it took me a while to clarify all the argument given. – Quang Hoang Dec 4 '18 at 4:33

Let $$A'$$ be the symmetric point of $$D$$ about $$A$$ on $$DA$$, i.e $$A'$$ is the point on $$DA$$ such that $$A'A = DA$$. Let $$F_1$$ be the intersection of $$DD_1$$ with $$AB$$ and $$F_2$$ be the intersection of $$DD_2$$ with $$AC$$. Then the following are easily observed:

(1) $$BF_1F_2C$$ is cyclic. This is because $$AF_1DF_2$$ is cyclic as $$\angle AF_1D = \angle AF_2D = 90^{\circ}$$. Then note that $$\angle BF_1F_2 = \angle BF_1D + \angle DF_1F_2 = 90^{\circ} + \angle DAF_2 = 90^{\circ} + 90^{\circ} - \angle C = 180^{\circ} - \angle BCF_2$$, proving the claim.

(2) The $$\triangle D_1A'D_2$$ is an expansion of $$\triangle F_1AF_2$$ with scaling factor $$2$$. So $$A'D_1$$ (resp. $$A'D_2$$) is parallel to $$AB$$ (resp $$AC$$) and hence $$E_1$$ (resp $$E_2$$) is the intersection of $$A'D_1$$ with $$BC$$ (resp. $$A'D_2$$ with $$BC$$). This shows that the quadrilateral $$D_1E_1E_2D_2$$ is obtained from the quadrilateral $$F_1BCF_2$$ by expansion about $$D$$ by a factor of $$2$$.

Since we've shown the cyclicity of $$F_1BCF_2$$ in (1), it follows that $$D_1E_1E_2D_2$$ is cyclic as well.

Haven't really thought of the circumcentrecentre as of now but should be tractable.

• I think you have a typo in $(1)$. Shouldn't $\angle BF_1C$ be $\angle BF_1D$? – Snip3r Dec 5 '18 at 19:29
• yes it is a typo, thank you for pointing out. – hellHound Dec 5 '18 at 19:30
• Also, can you please explain a little bit more about why $\angle DF_1F_2 = \angle DAF_2$? – Snip3r Dec 5 '18 at 19:32
• Because $AF_1DF_2$ is cyclic and these are angles subtended by the same chord. – hellHound Dec 5 '18 at 19:33