Proportionality and Similarity

One of my French tutees showed me the other day a question in her test she struggled with.

The figure represents the situation of a surveyor who wants to work out the height of a tree. The test was on what is called in France the “Théorème de Thalès” (Intercept Theorem in English), which states that two parallel lines cutting two intersecting lines creates segments of proportional lengths. In the standard situation shown in the figure, it means that

In the question shown above, however, we cannot apply this theorem directly as we don’t have the lengths FG and GE.

In the UK and USA, this topic is taught with similar figures and enlargement. Because we have two parallel lines, we easily recognise that ABC and EFC are similar triangles, which means that enlarging one of the triangles by the correct scale factor and about the correct point (here the point C) gives the other triangle. Knowing how enlargements work, looking at this situation from this perspective makes it clear that H is the image of G by this enlargement and so CH/CG = AB/EF.

However, for my French student, this is not obvious at all. So, using the theorem she knows, she could only conclude that CH/CG = BH/FG = AH/EG, while she wanted to find AB having EF, not BH and HA.

Now, we need to remember that this equality of quotients corresponds to a proportional relationship. If we translate it to a well-known example of a proportional relationship in our daily life, we could say for example that CH, BH and AH are the prices of CG, FG and EG kg of apples. The ratios CH/CG, BH/FG, and AH/EG are thus simply the price per kg of apples. Now, if I have bought 2 kg of apples for £6 and 3 kg for £9, I know that 2 + 3 = 5 kg cost £6 + £9 = £15, which is of course as well 5 kg x £3 per kg. So, intuitively, we will apply the rule that

This rule may look strange because it somehow reminds of one common misconception about adding fractions. But of course it has nothing to do with that here!

It can easily be proven with algebra. Since a’/a = b’/b = k, we can write that a’ = ka and b’ = kb, leading to a’+b’ = ka + kb = k (a+b). And so