103 Trigonometry Problems

188 103 Trigonometry Problems
We establish the following Lemma.
Lemma Let AD, BE, CF be the altitudes of acute triangle ABC, with
D, E, F on sides BC,CA,AB, respectively. Then
|DE|+|DF| ≤|BC|.
Equality holds if and only if |AB| =|AC|.
Proof: We consider Figure 5.13. Because ̸ CFA = ̸ CDA = 90 ◦ , quadrilateral
AFDC is cyclic, and so ̸ FDB = ̸ BAC = ̸ CAB and ̸ BFD =
̸ BCA = ̸ BCA. Hence triangles BDF and BAC are similar, so
or (by the double-angle formula)
|DF|
|AC| = |BF| = cos B,
|BC|
|DF| =b cos B = 2R sin B cos B = R sin 2B.
Likewise, |DE| =c cos C = R sin 2C. Thus,
|DE|+|DF| =R(sin 2B + sin 2C). (†)
Since 0 ◦