Four Concurrency Points In A Triangle

page revision: 0, last edited: 15 Sep 2008 09:01

Four Concurrency Points In A Triangle

page revision: 0, last edited: 15 Sep 2008 09:01

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Hi, how can I prove that extended altitudes of one triangle are angle bisectors of another triangle formed by intersections of this altitudes with a circumscribed circle?

ReplyOptionsI can tell you, but could you tell me first what connection you see with Exercise 323:

Into a given circle, inscribe a triangle such that the extensions of its altitudes intersect the circle at three given points?Connect three given points and draw angle bisectors - new intersection points of the bisectors and the circle are vertices of the triangle we are asked to construct.

Having seen ex. 325 I resolved my issue, as both problems have similar proofs.

Is there a simpler solution to ex. 323?

Frankly, I don't see much similarity between the arguments in 323 in 325. Are you sure your solution is correct?

But still my first question remains unanswered: in 323, what makes you think that the construction problem is solved by bisecting the arcs between the given 3 points on the circle? (It sounded as if you read the answer somewhere, but could not figure out why it works.)

ReplyOptionsYes, the solution seems correct. I would have posted an illustration if I could (low karma).

In 323 I proved by observing that two bisected angles intercept two arcs that are subtending two angles of the original triangle, which turn out to be the same, as they constitute right sub-triangles sharing one acute angle.

In 325 I used the same proof after circumscribing quadrilaterals formed by two vertices of the original triangle and two of the orthic one (capturing relevant angles).

When I first approached ex. 323, I constructed both triangles to see if there was any relation between them, and there apparently was. However I struggled to come up with a proof and that is what led me to the original question.

Oops, missed the thread.

ReplyOptionsIn 325, what you say sounds right, and 323 seems also right, if I am interpreting correctly what you are saying.

I am still puzzled though with your initial question about 323. As in any construction problem, it perfectly makes sense to assume that the required triangle has already been constructed and try to find a relationship, which would allow you to recover the required triangle from the given one. However, your saying that "there was a relation" means that you proved a relation, for - how else would you know that there was one?

I used GeoGebra for this problem. Once the figure was constructed, it allowed me to drag one of the vertices of the needs-to-be-constructed triangle and visually observe in dynamic what the second triangle looked like for all different combinations. When a possible relationship was identified, it was confirmed measuring the bisected angles instrumentally. So I knew it works but didn't know why.

Ah (so, my suspicion was almost correct), you discovered your conjecture "experimentally"!

By the way, given a triangle ABC, and the points A', B', C' of intersection of the extended altitudes with its circumcircle, if you conjecture that the arcs A'B and C'B are congruent, then you conclude that the inscribed angles BAA' and BCC' must be congruent, and then see that they are congruent indeed, since each combined with the angle ABC gives 90 degrees (this is the way

you proved your conjecture, I believe).

Alternatively, one can use the fact that each of the vertical angles between two chords is measured by the semisum of the arcs intercepted by these angles. The chords AA' and BC are perpendicular, and CC' and AB are perpendicular too. Thus the arc sums

BC' + AC and BA' + AC are the same, and so BC' and BA' are congruent.

Thank you!

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