In the given figure, a point O is taken inside an equilateral quadrilateral PQRS such that OP = OR. Show that Q, O and S lie on the same straight line.

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Solution

An equilateral quadrilateral means a quadrilateral whose all 4 side are equal.
Such a quadrilateral is Rhombus (special case is a square)
We take Rhombus which is a general case of quadrilateral
Let O be any point inside Rhombus such that OP = OR
Join PR. $Δ$ OPR is an isosceles triangle.
Draw OX perpendicular PR. This implies that OX is perpendicular bisector of PR $∵ Δ O P R$ is isosceles)
Now we also know that diagonals of a Rhombus bisect each oterh at right angles.
Hence, $S Q ⊥ P R$ and $S Q ⊥$ bisector of PR. But OX is also a $⊥$ bisector of PR.
$\Rightarrow$ OX lies on SQ $\Rightarrow$ Q, O, S lie in a straight line.

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Q. In the given figure, 'O' is a point inside a

Δ P Q R

such that

∠ P O R = 90^{\circ},

Op=6 cm and OR=8 cm. If PQ = 24 cm and

∠ Q P R = 90^{\circ},

then QR=

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