The triangle ABC has orthocentre H, so that HA is an antenna above the green hill. The rectangle BCDE is inscribed in the circumcircle (ABC). Show that HA = CD = BE.
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Antennas above and below
One reply on “Antennas above and below”
Let A, B, C denote (equal length) vectors from the center of the circle to those points. Then the vector sum A+B+C = H. (To show this we need to show that (H-A) is perpendicular to B-C, that H-C is perp to A-B, and that H-B is perp to A-C. But H-A = (A+B+C)-A = B+C, etc. So, we need to show that (A-B) is perp to A+B. But (A-B).(A+B) = A^2 – B^2 = |A|^2 – |B|^2 =0 because the vectors A, B, C all have the same length. Similarly, we show that (B-C).(B+C)=0 and (C-A).(C+A)=0.)
The antenna on the hill is represented by the vector H-A = B+C.
The vector of the “antenna in the ground” is B-E = B – (-C) = B+C.