Examples:
1) Given a triangle ABC, let P be a point on AB, let Q be a point on AC, and assume that PQ is parallel to BC. Prove that:

                           |AP|   |AQ|
                           ---- = ----
                           |PB|   |QC|
Solution:


Let x = |AP|, y = |PB|, u = |AQ|, and v = |QC| as shown below.  The triangles 
APQ and ABC are similar because the angle at A is common to both of them 
and APQ = ABC and AQP = ACB because PQ and BC are parallel.  Thus, if 
they are similar, their sides are in proportion.  That is:

 |AP|   |AQ|
 ---- = ----.
 |PB|   |QC|


That is, x/(x + y) = u/(u + v).  Cross-multiplying yields xu + xv = xu + yu.
Cancel xu term and obtain xv = yu.  If you divide both sides by v and also by
y, you find that x/y = u/v as was required.
2) A clever outdoorsman whose eye-level is 2 meters above the ground, wishes to find the height of a tree. He places a mirror horizontally on the ground 20 meters from the tree, and finds that if he stands at a point C which is 4 meters from the mirror B, he can see the reflection of the top of the tree. How high is the tree?

Solution:


We make the assumption that the man and the tree are both standing up 
straight and that the ground is flat.  So PBC = QBA also,
the triangles PCB and QAB are similar (by the proposition stated on the 
tutorial page).  Thus,   |QA|   |AB|
                         ---- = ----
                         |PC|   |CB|
                        
                         |QA|    20
                       = ---- = ----
                          2      4

                         |QA| = 10

Therefore, the height of the tree is 10 meters.
3) A child 1.2 meters tall is standing 11 meters away from a tall building. A spotlight on the ground is located 20 meters away from the building and shines on the wall. How tall is the child's shadow on the building?

Solution:

Let h be the height of the shadow on the building.  Then draw a diagram 
assuming the ground to be flat, as in the diagram below.




There are two triangles: one formed by the spotlight and the child, and one 
formed by the spotlight and the height of the shadow, h.  These two triangles 
share a common angle A at the spotlight.  If we assume that the child and the 
wall of the building are perpendicular to the ground, then the angle formed by 
the child and the ground (angle C) are both right angles.  So the triangles have 
another pair of equal angles.  Therefore, the trianlges are similar.

Now we must look at the lenghts of the corresponding sides.  We know that the 
child must be 9 meters from the spotlight (i.e. 20 m- 11 m).  This length in the 
smaller triangle corresponds to the distance from the spotlight to the building 
in the larger triangle (i.e. 20 m).  The height of the child in the smaller 
triangle (1.2 m) corresponds to the height of the shadow in the larger triangle 
(h).  Since the triangles are similar, these lengths are in proportion.  
Therefore:    9    1.2
             --- = ----
              20     h

             9h = 20 (1.2)      
              h = 24/9 = 8/3 = 2.67 meters

The height of the shadow is 8/3 meters (approx. 2.67 meters).

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