The scalene triangle ABC, has sides AB bigger than AC.
Folding the triangle along a line through A so that C folds onto the line AB, gives point P - the point where the line intersects the line AB.
I have the following ratio needing to be proved:
$$BA : AC = BP : PC$$
I have to prove this by looking at the areas of the triangle ABP and ACP in two different ways.
I believe I have found that:
- ABP = 2/3 of ABC, and ACP = 1/3
- ABP = 2 times ACP
But don't know how to use those facts to prove the ratio. How do I prove this?
Here is a quick (inaccurate) sketch (The dotted line represents the fold, and the grey line represents where APC ends up after folding):
Note: the triangle is folded back, and all calculations are done with the unfolded triangle.
The only hint I've been given is that I need to think about the heights of the triangles. Possibly rotating them onto a different base?
Drop $CD$ perpendicular to $AP$ at $D$, and $AE$ perpendicular to a line through $B$ parallel to $AP$. Thus $CD$ and $AE$ are the heights of triangles $ACP$ and $ABP$, respectively. Then since the areas of triangles $ABP$, $APC$ are proportional to bases $BP, PC$ [Euclid VI,1], and are also proportional to heights $AE,CD$, since they share base $AP$, we have $$\frac{BP}{PC}=\frac{AE}{CD}$$And since $\angle ABE = \angle BAP = \angle CAD$, therefore $\triangle ABE$ and $\triangle CAD$ are similar, making $$\frac{AE}{CD}=\frac{AB}{AC}$$Therefore,$$\frac{BP}{PC}=\frac{AB}{AC}$$