Cosines Rule
Three points A, B and C lie in this order on a line, and P is any
point in the plane. Use the Cosine Rule to prove the following
statement.
Age
14 to 16
Challenge level
(a) Prove that a triangle with sides $3$, $7$, $8$ contains a $60$ degree angle.
(b) Three points $A$, $B$ and $C$ lie in this order on a line, and $P$ is any point in the plane. Use the Cosine Rule to prove that:
\[{AP^2\over AB.AC}+{CP^2\over CA.CB} = 1 + {PB^2\over BA.BC}.\]
Image
Why is the $60$ degree angle opposite the side of median length?
For the second part all you need is the cosine formula and some algebraic manipulation to get the required result.
The Cosine Rule for $\Delta APC$ and $\Delta BPC$, where $\angle ACP=\theta$, gives $\begin{eqnarray} AP^2 &= AC^2+PC^2-2AC.PC \cos\theta,\cr PB^2&= BC^2+PC^2-2BC.PC \cos \theta. \end{eqnarray}$
Hence $\begin{eqnarray} \frac{BC^2+PC^2-PB^2}{ 2BC.PC}&= \frac{AC^2+PC^2-AP^2}{ 2AC.PC} = \cos\theta. \end{eqnarray}$
Hence, multiplying both sides by $2PC/AB$, we find that $\begin{eqnarray} {AP^2\over AC.AB} +{PC^2\over AB}\left({AC-BC\over BC.AC}\right) &= {PB^2\over AB.BC} +{AC-BC\over AB}.\end{eqnarray}$ As $AB+BC=AC$, we get the result: $\begin{eqnarray} {AP^2\over AB.AC}+{PC^2\over AC.BC} &= 1 + {PB^2\over AB.BC}. \end{eqnarray}$
Why do this problem?
The first part is an easy application of the cosine rule. The second part requires some algebraic skills and gives practice in proving a formula.
Possible approach
Use as practice. The problem does not require learners to decide on the method to use.
Key questions
Look at the formula to be proved. Write down the cosine formula for the triangles in the diagram using the same notation. Can you see how to manipulate the formulas you have established to give the formula to be proved?