6. Theorem of Ceva, Menelaus and Van Aubel.
6. Theorem of Ceva, Menelaus and Van Aubel.
6. Theorem of Ceva, Menelaus and Van Aubel.
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Figure 5:<br />
Multiplying the 3 ratios, we get concurrency <strong>of</strong> the altitudes. Furthermore,<br />
|HA|<br />
|HA2|<br />
= |C2A| |B2A|<br />
+<br />
|C2B| |B2C|<br />
= tan( B) + tan( C)<br />
tan( A)<br />
= sin( B + C). cos( A)<br />
cos( B) cos( C) sin( A)<br />
= sin(180◦ − A) cos( A)<br />
cos( B) cos( C) sin( A) =<br />
= tan( B)<br />
tan( A) + tan( C)<br />
tan( A)<br />
cos( A)<br />
cos( B) cos( C) .<br />
Lemma 1 Let ABC be a triangle <strong>and</strong> A1 a point on the side BC so<br />
that<br />
|A1B|<br />
|A1C|<br />
= γ<br />
β<br />
Let X <strong>and</strong> Y be points on the sides AB <strong>and</strong> AC respectively <strong>and</strong> let M be<br />
the point <strong>of</strong> intersection <strong>of</strong> the line segments XY <strong>and</strong> AA1 (Figure 6). Then<br />
β( |XB| C|<br />
) + γ(|Y ) = (β + γ)(|A1M|<br />
|XA| |Y A| |MA| ).<br />
Pro<strong>of</strong> First suppose that<br />
XY is parallel to the side BC. Then<br />
|XB|<br />
|XA|<br />
= |Y C|<br />
|Y A|<br />
= |MA1|<br />
|MA| ,<br />
5<br />
Figure 7:
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