Question 1031420
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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.
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We are given two circles with the centers O1 and O2 intersec-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
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, &nbsp;then the product of the measures 
of the secant and its external part is equal to the square 
of the tangent segment.
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{{{drawing( 400, 400, -3.5, 10.5, -5.5, 10.5, 
            circle(0.0, 0.0, 3, 3),

            circle(0.0, 0.0, 0.05, 0.05),
            locate (-0.1, -0.1, O1),

            circle(5.0, 0.0, 4, 4),

            circle(5.0, 0.0, 0.05, 0.05),
            locate (4.9, -0.1, O2),

            locate (2,  3.0, A),
            locate (2, -2.5, B),

        red(line(1.8, -3, 1.8, 10)),
            circle(1.8, 9, 0.08, 0.08),
            locate(2, 9.2, P),

       blue(line(1.8, 9, -3.65, 0)),
            circle(-2.5, 1.9, 0.05, 0.05),
            locate(-2.9, 2.3, Q),

       blue(line(1.8, 9, 11.2, 0)),
            circle(7.6, 3.5, 0.05, 0.05),
            locate(7.7, 3.9, R)
)}}}

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Regarding this theorem, see the lesson <A HREF=https://www.algebra.com/algebra/homework/Circles/Metric-relations-for-a-tangent-and-a-secant-lines-released-from-a-point-outside-a-circle.lesson>Metric relations for a tangent and a secant lines released from a point outside a circle</A> 
in this site.


Now, by applying this Theorem, we have &nbsp;&nbsp;{{{abs(PQ)^2}}} = |PA|*|PB|,  &nbsp;&nbsp;&nbsp;&nbsp;and &nbsp;&nbsp;&nbsp;&nbsp;{{{abs(PR)^2}}} = |PA|*|PB|.


Since the right sides are identical, the left sides are equal: &nbsp;&nbsp;{{{abs(PQ)^2}}} = {{{abs(PR)^2}}}. 


Then &nbsp;&nbsp;|PQ| = |PR|, &nbsp;&nbsp;QED.