Question

in triangle ABC,AB=AC P is the mid point of AC and Q os the mid point of AB prove that quadrilateral BCPQ is cyclic
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Answer 1

Solution :-

In ∆ABC we have given that,

→ AB = AC

So,

→ ∠ABC = ∠ACB = Let x. { Angle opposite to equal sides are equal in measure . } --------- Eqn.(1)

Join QP . since they are mid points of AB and AC .

→ QP || BC . { By mid point theorem. }

then,

→ ∠AQP = ∠ABC = x { corresponding angles. }

now, since A - Q - B is a straight line .

So,

→ ∠AQP + ∠BQP = 180° { Linear pair angles.}

→ x + ∠BQP = 180°

→ ∠BQP = (180° - x) ---------- Eqn.(2)

adding Eqn.(1) and Eqn.(2) we get,

→ ∠ACB + ∠BQP = x + (180° - x)

→ ∠PCB + ∠BQP = 180° .

similarly, we can conclude that,

→ ∠QBC + ∠QPC = 180° .

As we know that, in a cyclic quadrilateral, the sum of a pair of opposite angles is 180° .

therefore, we can conclude that, BCPQ is a cyclic quadrilateral .

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