Problem 16
Two circles, S and S’, intersect each other at P and Q. Any point A is taken on S, and joined to P and Q. AP and AQ are extended till B and C respectively, where B and C lie on S’. Show that:
BC is constant, as A varies
Here’s a simulation that demonstrates the problem.
Solution
Before you proceed with this problem, I suggest you head over to Problem 10 and try solving that one first. If you’ve already done that, then this problem shouldn’t require further thinking.
Anyways, here’s what you’ll need to solve the problem.
Similar Triangles
If two triangles have the same ‘shape’, then they are said to be similar. This means that the corresponding angles of these triangles are equal. Also, their corresponding sides are proportional.
For example, if ∆ABC and ∆DEF are similar, then
A = D, B = E, C = F
AB/DE = BC/EF = CA/FD
Here’s a simulation that shows two similar triangles. You can drag the vertices and observe the angles and side lengths.
To prove that two triangles are similar, you can prove that their corresponding angles as equal. Then, you can use the fact that their corresponding sides are proportional. Or, vice versa.
Cyclic Quadrilaterals
The opposite angles of a quadrilateral are supplementary. Here’s a simulation that demonstrates this property.
You can drag each vertex and observe the sum of the opposite angles. Is it always 180°?
Now, using the cyclic quadrilateral property, you can prove the similarity of the two triangles in the given figure. Then, using the proportionality of sides, can you prove that BC is of constant length?
Source: MathClub
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