Rebecca Parker
Assignment 9:

Proof that the orthocenter is the point of concurrency for the altitudes of a given triangle.  


 

    For a sketch of the given triangle and all the points, rays, and segments that I will be referring to, click here.

    We are constructed a segment GI is parallel to segment BC, and that segment HI is parallel to segment BA.Now, by definition of a parallelogram, we know that AGBC is a parallelogram.This implies also by the definition of a parallelogram that segment GA is congruent to segment BC, and that segment AC is congruent to BG by definition of parallelogram.


 
 

    In addition, given the fact that segment BA is parallel to segment HI and that segment GI is parallel to segment BC, we know that BCIA is a parallelogram by the definition of a parallelogram.This also implies that segment AI is congruent to segment BC and that segment AB is congruent to segment CI by definition of parallelogram.


 
 

    Now, because we have shown that segment BC is congruent to GA, and segment AI is congruent to BC, we can say that segment GA is congruent to AI by the reflexive property.We can also say that angle GAE and angle AEC are congruent because they are alternate angles cut by a transversal, and angles IAE and AEB are congruent for the same reason.Thus, angles AEC and AEB are equal to 90 degree because they are supplementary angles in a linear pair.Also, note that angles GAE and IAE are congruent by the reflexive property.
 
 

    Now that we have shown that the segments from the vertex to that base of the triangle (or the line parallel and on top of the base) are perpendicular, we know that segments AE, CD, and BF are the altitudes of the triangle, by definition of altitudes.  Also, note that:

        1)  AB is parallel to HI, so DC is also perpendicular to HI,
        2)  BC is parallel to IG, so AE must also be perpendicular to IG, and
        3)  AC parallel to GH, so BE is also perpendicular to GH.
1), 2), and 3) are all true because the altitudes of triangle ABC are all each the transversals that cut a set of parallel lines.  Thus, It holds by the properties of parallel line, that if the altitude is perpendicular to the base, that it will also be perpendicular to the outer triangle.  WE have already proven that segments AI and GA are congruent.  Thus, AE must be the perpendicular bisector of GI.  This similar property holds as you rotate the triangle.  Similarly, we can show that AE, CD, and BF are the perpendicular bisectors of the sides of the outer triangle.  Thus, these segments are concurrent at the circumcenter of triangle GIG.  Thus, all three of the altitudes will meet in a point at concurrency at the orthocenter of triangle ABC. 

 

Feel free to explore this investigation yourself by clicking here.


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