The Art and Craft of Problem Solving

8.3 SURVIVAL GEOMETRY II 275

8.3.9 Let ABC be a triangle and let m be a line parallel to BC, intersecting AB and AC

respectively at D and E. Then ADE "-' ABC.

Conversely, if points D and E lie on sides AB and AC, respectively, and divide

these sides proportionately (AD / AB = AE / AC), then DE II BC.

A

« mu--Esu o>

B C

8.3.10 Consider a right triangle with legs a and b and hypotenuse c. Drop a perpen­
dicular from the right angle to the hypotenuse, dividing it into segments with lengths
x andy.

c

(a) Prove that the two small triangles formed are similar, and similar to the large

triangle.

(b) Show that h = .;xy.

(c) Show that cx = a^2 and cy = b^2.

If a geometry problem involves ratios, it is almost certain that you will need to

investigate similar triangles. Look for them, and if they are not there at first glance, try

to produce them with auxiliary lines.

Solutions to the Three "Easy" Problems

With this strategy in mind, and the elementary review of geometry under your belt, you

should be able to solve the three "easy" problems (8.1.1-8.1 .3) given on pages 257-

258. Try them now, and check your solutions with the discussion below.

Example 8.3. 11 Proof of The Power of A Point Theorem (8. (^1) .1). The problem asks us
to prove an equality of products. This will involve either equating areas, or producing
similar triangles, equating ratios, and cross-multiplying. Let's try the second strategy.