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# Rita draws a right angled triangle. Let us prove logically that the perpendicular bisectors of sides are concurrent and where the position of the circum centre (inside/outside/on the side) is

Answer :

PO is perpendicular to AB and OQ is perpendicular to BC

Since Δ ABC is right angled at B so POQB forms a rectangle

Hence we can say that

PO = BQ (Opposite sides of rectangle)

BP = QO(Opposite sides of a rectangle)

In Δ AOP and Δ QOC

∠APO = ∠OQC (OP and OQ are perpendiculars)

∠OAP = ∠COQ (Corresponding angles)

PO = BQ = CQ (Q is midpoint of BC)

So Δ AOP and Δ COQ are congruent by A.A.S. axiom of congruency

Hence AO = OC

So O is the midpoint of AC

Hence the perpendiculars are concurrent

The circum centre lies on the midpoint of the hypotenuse of the triangle.

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