In the picture below, ΔPQR is right angled. Also, ∠A = ∠P and BC = QR.

Prove that the diameter of the circumcircle of ΔABC is equal to the length of PQ.

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Solution

Given: ∠PRQ = 90°

We know that angle in a semi circle is a right angle.

If we draw a circle passing through vertex R of ΔPRQ then PQ acts as the diametre of the circle.

We have ∠A = ∠P and BC = QR.

We know that angles in the same segment are equal.

BC and QR are equal segments having equal angles, ∠A and ∠P. So, points A, B and C also lie on the same circle on which points P, Q and R lie.

As PQ is the diametre of the circumcircle of ΔPQR, PQ is the diametre of the circumcircle of ΔABC.

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In the given figure, a circle is circumscribing $Δ A B C$ where $∠ A = 125^{\circ}$ and side BC=8cm. The diameter of the circumcircle is equal to

$[s i n 55^{\circ} = 0.82]$

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In the given figure, a circle is circumscribing $Δ A B C$ where $∠ A = 125^{\circ}$ and side BC=8cm. The diameter of the circumcircle is equal to

$[s i n 55^{\circ} = 0.82]$

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