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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):

enter image description here

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?

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scalene triangle 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,$$\frac{BP}{PC}=\frac{AB}{AC}$$

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  • $\begingroup$ AE? There isn't a point E.... $\endgroup$ – Beastly Gerbil Apr 17 '17 at 19:12
  • $\begingroup$ Through B draw a line parallel to AP. From A drop a perpendicular to that line, calling the point where they meet E. $\endgroup$ – Edward Porcella Apr 17 '17 at 20:20
  • $\begingroup$ Does this edit clarify the answer? Euclid VI,3 proves the proportion without considering areas, but I think the above makes full use of the hints you mention: look at areas, think about heights, rotate onto different base. What do you think? $\endgroup$ – Edward Porcella Apr 18 '17 at 17:57

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