# Orthogonality in Isosceles Triangles

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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.

Discussion

### 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.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 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 CP/PD = CE/ES = 2. DF||BC as a midline.

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, DO ⊥ PE, for PE||AB. Also, PO ⊥ DE because, DE||BC, DF being the midline. O is therefore the orthocenter of triangle PDE. It follows that EO is the third altitude and is perpendicular to the third side, DP.

Conversely, if EO ⊥ DP (CD), then since OD ⊥ PE, again O is the orthocenter of PDE. Hence, the altitude from P passes through O, such that in triangle ABC the circumcenter O and the centroid P are collinear with the midpoint of BC. Thus the median from A is also the altitude and ABC is bound to be isosceles.

### Solution 2

Place the origin at O and let the points be identified with their radius-vectors. Then D = (A + B)/2 and E = (D + A + C)/3 = (3A + 2C + B)/6. EO ⊥ CD is equivalent to

 (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,

 (3A + 2C + B)·(A + B - 2C) = 3A2 + 2A·C + A·B + 3A·B + 2B·C + B2 - 6A·C - 4C2 - 2B·C = 4A·B - 4A·C = 4A·(B - C).

Also,

 (A - C)·(A - C) = (A - B)·(A - B) ≡  - 2A·C + C2 = - 2A·B + B2 ≡  - 2A·C = - 2A·B ≡  2A·(B - C) = 0.

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

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 OE ⊥ FS iff AD = AF.

Why does this remind me of another problem?

### References

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