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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Then<br />
Pro<strong>of</strong> Let x = |AX| <strong>and</strong> y = |AY |.<br />
x + y =<br />
a + b + c<br />
2<br />
where a, b <strong>and</strong> c are lengths <strong>of</strong> sides.<br />
Furthermore,<br />
so<br />
1<br />
2<br />
. . . (a)<br />
= area(AXY )<br />
area(ABC) = xy sin( A)<br />
bc sin( A) ,<br />
xy = bc<br />
2<br />
. . . (b).<br />
Consider b( |XB| C|<br />
) + c(|Y<br />
|XA| |Y A| )<br />
c − x − y<br />
= b( ) + c(b )<br />
x y<br />
= b( 1 1<br />
+ ) − b − c<br />
x y<br />
a + b + c<br />
= bc( .<br />
2<br />
2<br />
) − b − c<br />
bc<br />
= a.<br />
Thus by Corollary 2, incentre I belongs to the line XY.<br />
Figure 9:<br />
<strong>Theorem</strong> 6 Let ABC be an equilateral triangle <strong>and</strong> X, Y <strong>and</strong> Z points<br />
on the sides BC, CA <strong>and</strong> AB respectively (Figure 10). Then the minimum<br />
value <strong>of</strong><br />
|ZX| 2 + |XY | 2 + |Y Z| 2<br />
is attained when X, Y, Z are the midpoints <strong>of</strong> the sides.<br />
9