HamiltonHardison’s Exploration of

Assignment 8: An Altitude and Area Proof

From Jim Wilson’s website: Construct any acute triangle ABC and its circumcircle. Construct the three altitudes AD, BE, and CF. Extend each altitude to its intersection with the circumcircle at corresponding points P, Q, and R.

Find            .

Click HERE for a GSP sketch.

Upon investigation it appears that .

It may be helpful to add some line segments.  In order to find similar and congruent triangles.

Here we can see that  by angle-angle similarity (Red and Blue) (Apologies if colors do not show up on some computers).  Since BE is an altitude, HEC is a right angle.  Since CF is an altitude, HFB is a right angle.  Angles EHC and FHB are vertical angles.  We also have  by angle-angle similarity (Red and Yellow).  Since Arc AR is subtended by angles HCE and RBF, these angles are congruent.  Since BRF is the supplement of a right angle, it is also right.  We already know HEC is a right angle.

Since  and , .  Since FB is congruent to itself, we have that the blue and yellow triangles are congruent, or .  A similar argument shows that .  Now we have the two larger triangles, .

From similar reasoning we have:           ;  ;


The area of quadrilateral .  The area of quadrilateral .  The area of  quadrilateral

By triangle congruence, the sum of the blue areas (triangles BPC, CQA, and ARB) is equal to the area of the triangle ABC.  By adding the area of each of the quadrilaterals described above, we are computing 4-times the area of triangle ABC, (3 yellow triangles ABC and 3 blue triangles equal in area to ABC).

.  For convenience, we will write:


We can compute the area of ABC by using the altitudes as well.  .

Now we can combine our equations cleverly.

.  Since AB is congruent to itself, we have  as desired.


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