43
Theorem 6 (Stewart Theorem)
Let P be an arbitrary point in the side BC of \(\bigtriangleup\) ABC. If
the point P is distinct from the points A &C, then \(\text{AB}^{2}\)PC
+ \(\text{AC}^{2}\)BP =\(\text{AP}^{2}\ \bullet\) BC +
BP\(\ \bullet\) BP\(\ \bullet\) BC (1) or
. (2)
For the sake of convenience, we let a, b, c be the lengths of the sides
of the \(\bigtriangleup\) ABC which correspond to ∠ A, ∠ B, ∠ C in
side BC which divide a into two segments of lengths m and n with
rearrangement, we can express the above expression by man + dad = bmb +
cnc. “ A man and his dad put a bomb into the sink”.
Proof: Without loss of generality, we assume ∠ APC < 90\(\ \).
Then by using cosine law, we have
\begin{equation}
\text{AB}^{2}=AP^{2}+BP^{2}-2AP\bullet BP\bullet cos(180-\angle\text{APC})\nonumber \\
\end{equation}
\begin{equation}
{=AP}^{2}+BP^{2}+2AP\bullet BP\bullet cos\angle\text{APC}\nonumber \\
\end{equation}
Diagram 22
Multiple BP, PC into (1) and (2) and adding up. Then we have proved the
theorem.
As the converse part of Stewart theorem, we let B, P, C be the points in
the projective lines AB, AP, AC. Then we claim that if
, then B, P, C are collinear.
44
Corollary 6.5 (Stewart theorem)
(i) If P is a point in the extension line of BC, then
.
(ii) If P is a point in the opposite extension line of BC, then
.
Corollary 6.6
-
If \(\bigtriangleup\)ABC is an isosceles triangle and P is in BC, then
\(AP^{2}=AB^{2}-BP\bullet PC\)
-
If AP is a median of the side BC, then
\(AP^{2}=\frac{1}{2}AB^{2}+\frac{1}{2}AC^{2}-\frac{1}{4}BC^{2}\)
.
-
IF AP is the inner angle \(\angle\)A Bi, then
\(AP^{2}=AB\bullet AC-BP\bullet PC\)
-
If AP is the exterior angle \(\angle\)A, then
\(AP^{2}=-AB\bullet AC+BP\bullet PC\)
-
If P divides BC in
\(\frac{\text{BP}}{\text{BC\ }}=\lambda,\ then\ AP^{2}=\lambda\left(\lambda-1\right)BC^{2}+\left(1-\lambda\right)AB^{2}+\lambda AC^{2}.\)
-
If
\(\frac{\text{BP}}{\text{PC}}=k,\ then\ AP^{2}=\frac{1}{1+K}\bullet AB^{2}+\frac{k}{1+k}AC^{2}-\frac{k}{\left(1+k\right)2}\bullet BC^{2}.\)
We give some examples to show the application of Stewart theorem
Example 6.7 (high school students MO 1996, Beijing)
In the convex quadreteral ABCD,
at O. Find \(\angle\) AOB.
Solution
Extend BA, CD to meet at P. Let BC=x, then PB=2x, PC=\(\sqrt{3}x\). In
have
\begin{equation}
CA^{2}=PC^{2}\bullet\frac{\text{AB}}{\text{PB}}+BC^{2}\frac{\text{PA}}{\text{PB}}-AB\bullet PA\nonumber \\
\end{equation}
\begin{equation}
=\left(\sqrt{3}X\right)^{2}\bullet\frac{2}{2X}+x^{2}\bullet\frac{2x-2}{2x}-2(2x-2)\nonumber \\
\end{equation}
=\(x^{2}-2x+4\)
Diagram 23
45
Now, by right angled \(\bigtriangleup\)ADP ~ right
angled \(\bigtriangleup\)CBP, we have
.
Therefore,
A summer camp problem 2001 in China
In the following diagram, ove that
Proof
Diagram 24
Because EF and PB meet at C
We have \(\text{EC}\bullet CF=AC\bullet CB.\)
Because PE=PF, apply Stewart theorem corollary 6.6 (a), we have
\begin{equation}
=PC^{2}+\left(PC-PA\right)\bullet\left(PB-PC\right)\nonumber \\
\end{equation}
\begin{equation}
=PC^{2}-PC^{2}-PA\bullet PB+PC\bullet PB+PC\bullet PA\nonumber \\
\end{equation}
\begin{equation}
\text{Hence\ P}E^{2}=PA\bullet PB,\ consequently,\ 2PA\bullet PB=PA\bullet PC+PB\bullet PC\nonumber \\
\end{equation}
\begin{equation}
\text{Therefore\ }\frac{2}{\text{PC}}=\frac{1}{\text{PA}}+\frac{1}{\text{PB}}\text{\ .}\nonumber \\
\end{equation}
The proof is completed.
46
§ 7 Erdös-Mordell inequality
We first state the following theorem.
distance from the point P to the three sides be \(|PD|\) , \(|PE|\) and
+\(\ |PE|\) +\(\ |PF|\) .
Proof : We use polar coordinates to prove this theorem. Let P be the
polar point, \(|PA|\) is denoted by polar coordinates. Then
Write \(|PQ|\) =t1, \(|PR|\) =t2, \(|PS|\) =t3, the polar coordinates of
the point \(\theta\), R, S are
Because A, \(\theta\), B are collinear, by Ceva theorem, we have
\begin{equation}
\frac{\sin{(0-2\theta_{1})}}{t_{1}}+\frac{\sin{(2\theta_{1}-\theta_{1})}}{\varphi_{1}}+\frac{\sin{(\theta_{1}-o)}}{\varphi_{2}}=0\nonumber \\
\end{equation}
\begin{equation}
\text{Hence\ }t_{1}=\frac{2\varphi_{1}\varphi_{2}}{t_{1}}+\frac{\sin\left(2\theta_{1}-\theta\right)}{\varphi_{1}}+\frac{\sin\left(\theta_{1}-0\right)}{\varphi_{2}}=0\nonumber \\
\end{equation}
\begin{equation}
Therefore,\ t_{1}=\frac{2\varphi_{1}\varphi_{2}}{\varphi_{1+\varphi_{2}}}\cos\theta_{1}\leq\sqrt{\varphi_{1}\varphi_{2}}\cos\theta_{1}\nonumber \\
\end{equation}
\begin{equation}
Similarly\ we\ have\ B\ (\varphi_{2},2\theta_{1})\nonumber \\
\end{equation}
\begin{equation}
t_{2}\leq\sqrt{\varphi_{2}\varphi_{3}}\cos{\theta_{2},\ t_{3}\leq\sqrt{\varphi_{3}\varphi_{1}\cos\theta_{3}}}\nonumber \\
\end{equation}
Diagram 25
47 part 1
By using the known triangle inequality if
Since we have shown that
we can immediately verify that
We can also verify that the equality holds ot and only if \(ABC\ \)is an
equilateral triangle and P is its centroid.
Many inequalities can be removed by using points and lines of a triangle
as inspired by the proof of Erdös-Mordell inequality.
Example 7.2
Suppose that a, b, c, x, y, z are positive numbers satisfying the
condition \(a+x=b+y=c+z=k,\) prove that
Proof : Construct an equilateral triangle PQK with the length if the
side be k. In the three sides of the triangle, let the points N, M, L be
such that QL = x, LK= a, KM = y, MP = b, PN= z, NQ= c. Then it is clear
that \(S_{LKM}+S_{MPN}+S_{NRL}<S_{PQK}\)
DIAGRAM 26
Therefore
\begin{equation}
\frac{\sqrt{3}}{4}ay+\frac{\sqrt{3}}{4}bz+\frac{\sqrt{3}}{4}cx<\frac{\sqrt{3}}{4}k^{2}\nonumber \\
\end{equation}
Thus, \(ay+bz+cx<k^{2}\)
47 part 2
Erdös-Mordell inequality
Erdös Mordell inequality can also be proved by using observation
diagram.
Consider the diagram: