Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

## Friday, May 1, 2020

### Dynamic Geometry 1477: Miquel's Pentagram Theorem, Pentagon, Triangle, Circumcircles, Concyclic Points, Step-by-step Illustration

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Consider Triangle A1ED1, B1DC and C1AB are transversals.

ReplyDeletePoints D2 and A2 will also lie on circumcircle of triangle A1ED1.

Angle EA2D2= Angle EA1D2= Angle DC2D2= x,

Let Angle E2A2E=y. Angle E2A2D2=Angle EA2D2+Angle EA2E2=x+y

Now consider triangle E1C1D, AEB1 and BCA1 are transversal.

Point C2 and E2 lie on circumcircle of triangle E1C1D.

Angle E2C2D = Angle E2C1D = Angle E2A2E = y

Angle E2C2D2= Angle E2C2D + Angle DC2D2= x+y

We have Angle E2A2D2 = Angle E2C2D2, hence A2,C2,D2,E2 are concyclic.

Similarly B2 can be proven concyclic.

https://photos.app.goo.gl/QCqGuc6qFkjEkbWG7

ReplyDeletePradyumna’s solution is very good

I would like to add some clarification to his solution.

Consider circumcircles of triangles BDD1 and CDA1 and 2 secants BDA1 and CDB1 . these secants meet at D1

Per the result of problem 948 ( see link above)

A1, B1, D1 and D2 are cocyclic ( see blue circle)

Similar result to circumcircles of triangles AEC1 and ABD1 and 2 secants BAC1 and EAD1 we will have A2 on blue circles

Thanks Peter for your kind words.

DeleteAlso thanks for sharing nice diagram, it provides more clarity.

Regards

Pradyumna