Sprdalo M.

asked • 10/29/19# Help with geometry problem

~~Let $ABC$ be a triangle. Points $A_1, B_1, C_1$ are, respectively, on sides $BC, AC, AB$ and:~~

~~$\frac{AC_1}{C_1B} = \frac{BA_1}{A_1C} = \frac{CB_1}{B_1A} \neq 1$~~

~~If $\frac{AB}{A_1B_1} = \frac{BC}{B_1C_1} = \frac{CA}{C_1A_1}$ prove that triangle $ABC$ is equilateral.~~

I found this problem in the section where problems are based on complex numbers, complex geometry, polynomials,... However, any solution would be welcomed (especially complex one :) ) .Thanks in advance.

In case Latex is not working:

Let ABC be a triangle. Points A1, B1, C1 are, respectively, on sides BC, AC, AB and:

AC1/C1B} = BA1/A1C = CB1/B1A which is not equal to 1

If AB/A1B1 = BC/B1C1} = CA/C1A1 prove that triangle ABC is equilateral.

## 2 Answers By Expert Tutors

I was working on this problem also before Sam Z answered it.

I have not solved it all yet, but I do know enough to disagree with Sam Z.

What the first inequality tells you is that A1, B1, and C1 are NOT the midpoints of the sides.

What the equality tells you is that ΔABC is similar to ΔA1B1C1 because the sides are in proportion.

OK, I got it now.

The 3 triangles AB1C1, BC1A1, and CA1B1 are also similar because the sides are in proportion.

That means that ∠A = ∠B = ∠C because they are corresponding angles of similar triangles! QED

Sprdalo M.

Can you please explain what sides are in proportion? I don't quite seem to understand why these three triangles are similar...10/30/19

Paul M.

10/30/19

Paul M.

11/05/19

Sam Z. answered • 10/29/19

Math/Science Tutor

Without a display; I figure the 1s are = midpoints. This equilateral is divided into quarters. 3 angles and the center triangles.

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Mark M.

"AC1/C1B} = BA1/A1C = CB1/B1A != 1" is not a well defined equation. Please revise.10/29/19