# Orthogonality in Isosceles Triangles

What if applet does not run? |

In triangle ABC, O is the circumcenter, D is the midpoint of AB, E is the centroid of triangle ACD. Prove that EO is perpendicular to CD iff AB = AC.

|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener|

Copyright © 1996-2018 Alexander Bogomolny

### Discussion

The applet was supposed to remind of the following problem:

In triangle ABC, O is the circumcenter, D is the midpoint of AB, E is the centroid of triangle ACD. Prove that EO is perpendicular to CD iff AB = AC.

The problem has been drawn from [Honsberger, pp. 230-234]. Two solutions (trigonometric and vector+trigonomentry) are found there and a mention that additional two solutions appeared in Crux Mathematicorum, 1991, 105 and 228. The solutions below have been adapted from a discussion at the AoPS site.

What if applet does not run? |

Let M denote the midpoint of BC, P the centroid of ΔABC, S the midpoint of AD, F the midpoint of AC.

### Solution 1

PE||AB because P is on a median CD of ΔABC while E is on the median of ΔACD through C, and also

If AB = AC, AM serves as the altitude and the median of ΔABC from A. Thus both P and O belong to AM. In ΔPDE,

Conversely, if EO ⊥ DP (CD), then since

### Solution 2

Place the origin at O and let the points be identified with their radius-vectors. Then

(1) | (3A + 2C + B)·(A + B - 2C) = 0, |

where the dot stands for the scalar product. On the other hand, AB = AC is the same as |A - C| = |A - B| and, in terms of the scalar product,

(2) | (A - C)·(A - C) = (A - B)·(A - B). |

Because of our selection of the origin, A·A = B·B = C·C. We have,

^{2} + 2A·C + A·B + 3A·B + 2B·C + B^{2} - 6A·C - 4C^{2} - 2B·C | |

Also,

(A - C)·(A - C) = (A - B)·(A - B) | ≡ - 2A·C + C^{2} = - 2A·B + B^{2} |

≡ - 2A·C = - 2A·B | |

≡ 2A·(B - C) = 0. |

Therefore, (1) and (2) are either both true or both false. Both are equivalent to

It appears the problem has an equivalent formulation:

Let ADOF be a cyclic quadrilateral with AO a diameter. Let E on DF trisect DF: DE = 2EF, and let S be the midpoint of AD. Then

Why does this remind me of another problem?

### References

- R. Honsberger,
*From Erdös To Kiev*, MAA, 1996, pp. 230-234.

|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener|

Copyright © 1996-2018 Alexander Bogomolny