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Question
In the given figure, BC is a tangent to the circle with centre O. OE bisects AP. Prove that ΔAEO~Δ ABC.
Solution
The figure given in the question is below
Let us first take up ΔAOP.
We have,
OA = OP (Since they are the radii of the same circle)
Therefore, ΔAOP is an isosceles triangle. From the property of isosceles triangle,
we know that, when a median drawn to the unequal side of the triangle will be
perpendicular to the unequal side. Therefore,
∠OEA=90°
Now let us take up ΔAOE and ΔABC.
We know that the radius of the circle will always be perpendicular to the tangent at
the point of contact. In this problem, OB is the radius and BC is the tangent
and B is the point of contact. Therefore,
∠ABC=90°
Also, from the property of isosceles triangle we have found that
∠OEA=90°
Therefore,
∠ABC=∠OEA
∠A is the common angle to both the triangles.
Therefore, from AA postulate of similar triangles,
ΔAOE~ΔABC
Thus we have proved
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