SOLUTION: Hello I want to prove this T.T but i dont know how I need to draw this diagram and prove it! Two circles intersect at points A and B. From any point P on the line AB, tangents P

Algebra ->  Geometry-proofs -> SOLUTION: Hello I want to prove this T.T but i dont know how I need to draw this diagram and prove it! Two circles intersect at points A and B. From any point P on the line AB, tangents P      Log On


   



Question 1031420: Hello I want to prove this T.T but i dont know how I need to draw this diagram and prove it!
Two circles intersect at points A and B. From any point P on the line AB, tangents PQ and PR are drawn to the circles. Prove that PQ=PR.
Thank you!

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
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Hello I want to prove this T.T but i dont know how I need to draw this diagram and prove it!
Two circles intersect at points A and B. From any point P on the line AB, tangents PQ and PR
are drawn to the circles. Prove that PQ=PR.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

We are given two circles with the centers O1 and O2 intersec-        
ting in the common points A and B (see the Figure on the right).
The straight line AB is drawn through the points A and B.
The point P is an arbitrary point in this straight line.
The tangent lines PQ and PR are drawn from P to the circles O1
and O2 respectively. Q and P are the tangent points.
We need to prove that the tangent segments PQ and PR are
congruent: |PQ| = |PR|.

The key for the proof is this Theorem of the school Geometry:

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If a tangent and a secant lines are released from a point
outside a circle,  then the product of the measures
of the secant and its external part is equal to the square
of the tangent segment.
------------------------------------------------



Regarding this theorem, see the lesson Metric relations for a tangent and a secant lines released from a point outside a circle
in this site.

Now, by applying this Theorem, we have   abs%28PQ%29%5E2 = |PA|*|PB|,     and     abs%28PR%29%5E2 = |PA|*|PB|.

Since the right sides are identical, the left sides are equal:   abs%28PQ%29%5E2 = abs%28PR%29%5E2.

Then   |PQ| = |PR|,   QED.