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Let A , B and C be 3 points of a circle (c)
P is the intersection of two tangents of the circle in points B and C
Let (AB)//(CP) and AB=3 and BP=4 What is the length of BC
Can someone give hint ! Im blocked , the problem is that i don't know to start my proof
I see that BPC is triangle such that CP=BP
We suppose that ABPC is quadrilater parallelogram
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$\begingroup$It is not difficult to show that XBPC is a parallelogram. In fact, it is a rhombus because $PB = BX = 4.$
$\alpha = \beta$ (tangent properties)
$\beta = \gamma$ (alternate angles)
$\beta = \delta$ (angles in alternate segment)
$∴ \triangle PBC \sim \triangle CBA$
$∴ \dfrac {PB}{BC} = \dfrac {BC}{BA}$
That is, $\dfrac {4}{BC} = \dfrac {BC}{3}$
Result follows.
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