In - ABC - AB - AC - P is the mid point of AB and Q is the mid point of AC then-A- PQBC is a parallelogramB- BPCQ is a cyclic quadrilateralC- PQBC is a squareD- BPCQ is a rectangle

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Properties of Cyclic Quadrilateral

In Δ ABC , AB...

Question

In $Δ$ ABC, AB= AC. P is the mid point of AB and Q is the mid point of AC then:-

PQBC is a parallelogram

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BPCQ is a cyclic quadrilateral

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PQBC is a square

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BPCQ is a rectangle

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Solution

The correct option is B
BPCQ is a cyclic quadrilateral

AB= AC

Therefore $∠$ B= $∠$ C

PQ||BC (mid point theorem)

$∴ ∠$ C+ $∠$ Q= $180^{o}$ (CO-interior angles)

$∠$ B+ $∠$ Q= $180^{o}$ (Since $∠$ B= $∠$ C)

Hence, BPCQ is a cyclic quadrilateral

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In Δ ABC, AB= AC. P is the mid point of AB and Q is the mid point of AC then:-

The correct option is B BPCQ is a cyclic quadrilateral AB= AC Therefore ∠B= ∠C PQ||BC (mid point theorem) ∴∠C+∠Q= 180o (CO-interior angles) ∠B+ ∠Q= 180o (Since ∠B= ∠C) Hence, BPCQ is a cyclic quadrilateral

In $Δ$ ABC, AB= AC. P is the mid point of AB and Q is the mid point of AC then:-

The correct option is B
BPCQ is a cyclic quadrilateral

AB= AC

Therefore $∠$ B= $∠$ C

PQ||BC (mid point theorem)

$∴ ∠$ C+ $∠$ Q= $180^{o}$ (CO-interior angles)

$∠$ B+ $∠$ Q= $180^{o}$ (Since $∠$ B= $∠$ C)

Hence, BPCQ is a cyclic quadrilateral