Easy Maff Question

[Peter Pedant]

ABCD is a trapesium with AB parallel to DC - the diagnonals meet at X - prove DXC and BXA are similar traingles and that

area of DXC = DC2
---------------- ------
araa of AXB AB2

Answer - the similar traingles is easy - Z angles - F angles and corresponding angles
so there fore sinilar

you can say that DC /AB is a ratio let us say 'r' - but how do you show that the altitudes which you need for area are also in the ration 'r' ? - and then the areas are in the ratio r squared which is also DC/AB squared.... ergo - or do you just say - "they are"

The only thing I could manage was
imagine two similar triangles with side ratio 'r' but the altitude NOT r -
and let r approach one - the triangles are now congruent but their areas are not equal .....
which is impossible therefore the ratio of altitudes must also be 'r'

is there a simpler way?

(engtrance exam to secondary school)

[Rev. Green]

At this level, I'd give full marks to someone saying that these are corresponding lengths in similar shapes and area scales as the square of length - no calculation required - and a bonus mark for saying that the question can't be answered without more information.
The answer to a question depends on the exam. At university, one question was to show that closed is the opposite of open. After three years of obfuscation it is possible to take two pages to answer this. Luckily, a question on the previous year's paper had asked to show that open is the opposite of closed.