103 Trigonometry Problems

176 103 Trigonometry Problems
Third Solution: Because the circle ω is inscribed in ABCD, as shown in
Figure 5.12, we can set ̸ DAI = ̸ IAB = a, ̸ ABI = ̸ IBC = b, ̸ BCI =
̸ ICD = c, ̸ CDI = ̸ IDA = d, and a + b + c + d = 180 ◦ . Our proof is
based on the following key Lemma.
Lemma: If a circle ω, centered at I, is inscribed in a quadrilateral ABCD,
then
|BI| 2 + |AI| ·|BI|·|CI|=|AB|·|BC|. (†)
|DI|
a
b
A
d
a a
d
D
d
B
c
I
b
b
Figure 5.12.
c
c
C
Proof: Construct a point P outside of the quadrilateral such that triangle
ABP is similar to triangle DCI. We obtain
̸ PAI + ̸ PBI = ̸ PAB + ̸ BAI + ̸ PBA+ ̸ ABI
= ̸ IDC + a + ̸ ICD + b
= a + b + c + d = 180 ◦ ,
implying that the quadrilateral PAIB is cyclic. By Ptolemy’s theorem, we
have |AI|·|BP|+|BI|·|AP |=|AB|·|IP|, or
|BP|· |AI| +|BI|·|AP | =|AB|. (††)
|IP| |IP|
Because PAIB is cyclic, it is not difficult to see that, as indicated in the
figure, ̸ IPB = ̸ IAB = a, ̸ AP I = ̸ ABI = b, ̸ AIP = ̸ ABP = c,
and ̸ PIB = ̸ PAB = d. Note that triangles AIP and ICB are similar,
implying that
|AI|
|IP| = |IC| |AP |
and
|CB| |IP| = |IB|
|CB| .
Substituting the above equalities into the identity (††), we arrive at
|BP|· |CI|
|BC| + |BI|2
|BC| =|AB|,