Difference between revisions of "Symmedians, Lemoine point"
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<cmath>\angle BDA = \angle ACB = \angle ACM \implies \triangle ABD \sim \triangle AMC \implies</cmath> | <cmath>\angle BDA = \angle ACB = \angle ACM \implies \triangle ABD \sim \triangle AMC \implies</cmath> | ||
<cmath>\frac {AM}{MC}= \frac {AB}{BD}.</cmath> | <cmath>\frac {AM}{MC}= \frac {AB}{BD}.</cmath> | ||
− | Similarly <math>\triangle | + | Similarly <math>\triangle ABM \sim \triangle ADC \implies \frac {AM}{MB}= \frac {AC}{CD}.</math> |
<cmath>BM = MC \implies \frac {BD}{CD} = \frac{AB}{AC}.</cmath> | <cmath>BM = MC \implies \frac {BD}{CD} = \frac{AB}{AC}.</cmath> | ||
Revision as of 11:47, 9 July 2024
The reflecting of the median over the corresponding angle bisector is the symmedian. The angle formed by the symmedian and the angle bisector has the same measure as the angle between the median and the angle bisector, but it is on the other side of the angle bisector. The symmedian is isogonally conjugate to the median
There are three symmedians. They are meet at a triangle center called the Lemoine point.
Proportions
Let be given.
Let be the median,
Prove that iff is the symmedian than
Proof
1. Let be the symmedian. So
Similarly
By applying the Law of Sines we get
Similarly,
2.
As point moves along the fixed arc
from
to
, the function
monotonically increases from zero to infinity. This means that there is exactly one point at which the condition is satisfied. In this case, point
lies on the symmedian.
Similarly for point
Corollary
Let be the
symmedian of
Then is the
symmedian of
is the
symmedian of
is the
symmedian of
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