Question

In the adjoining figure, the bisectors of ∠B and ∠D of a quadrilateral ABCD meet CD and AB produced at P and Q respectively. prove that​

Answer 1

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$\huge{\underline{\underline{\sf{\orange{</p><p>Question:-}}}}}$

In the adjoining figure, the bisectors of angle B and angle D of a quadrilateral ABCD meet CD and AB produced at P and Q respectively. prove that

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$\huge{\underline{\underline{\sf{\pink{Solution:-}}}}}$

In ∠ADQ,we have

$\angle \: A \: + \angle \: ADQ + \angle \: Q = 180\degree \:$

$⇒ \angle \: A + \frac{1}{2} \angle \: D + \angle \: Q = 180\degree$

In ∆CBP,we have

$\angle \: C + \angle \: CBP + \angle \: P = 180\degree$

$⇒ \angle \: C + \frac{1}{2} \angle \: B + \angle \: P = 180\degree$

Adding (i) and (ii),we get

$\angle \: A + \angle \: C + \frac{1}{2} \angle \: B + \frac{1}{2} \angle \: D + \angle \: P + \angle \: Q = 360\degree$

.

$⇒ \angle \: A +\angle \: C + \angle \: B + \angle \: D + \angle \: P + \angle \: Q = 360\degree + \frac{1}{2} \angle \: B + \frac{1}{2} \angle \: D \\ \: \: \: \: \: \: \: (\therefore \: adding \: ( \frac{1}{2} \angle \: B + \frac{1}{2} \angle \: D) \: on \: both \: sides)$

$⇒ 360\degree + \angle \: p + \angle \: Q = 360\degree + \frac{1}{2} (\angle \: B + \angle \: D) \\ \: \: \: \: \: \: \: \: (\therefore \: sum \: of \: all \: the \: angles \: of \: a \: quadrilateral \: is \: 360\degree)$

$⇒ \angle \: P + \angle \: Q = \frac{1}{2} (\angle \: B + \angle \: D).$

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Answer 2

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question - In the adjoining figure, the bisectors of angle B and angle D of a quadrilateral ABCD meet CD and AB produced at P and Q respectively. prove that..

Solution -

$In ∠ADQ,we have</p><p></p><p>\angle \: A \: + \angle \: ADQ + \angle \: Q = 180\degree \:∠A+∠ADQ+∠Q=180°</p><p></p><p>⇒ \angle \: A + \frac{1}{2} \angle \: D + \angle \: Q = 180\degree⇒∠A+21∠D+∠Q=180°</p><p></p><p>In ∆CBP,we have</p><p></p><p>\angle \: C + \angle \: CBP + \angle \: P = 180\degree∠C+∠CBP+∠P=180°</p><p></p><p>\begin{gathered}⇒ \angle \: C + \frac{1}{2} \angle \: B + \angle \: P = 180\degree \\\end{gathered}⇒∠C+21∠B+∠P=180°</p><p></p><p>Adding (i) and (ii),we get</p><p></p><p>\begin{gathered}\angle \: A + \angle \: C + \frac{1}{2} \angle \: B + \frac{1}{2} \angle \: D + \angle \: P + \angle \: Q = 360\degree \\\end{gathered}∠A+∠C+21∠B+21∠D+∠P+∠Q=360°</p><p></p><p>.</p><p></p><p>\begin{gathered}⇒ \angle \: A +\angle \: C + \angle \: B + \angle \: D + \angle \: P + \angle \: Q = 360\degree + \frac{1}{2} \angle \: B + \frac{1}{2} \angle \: D \\ \: \: \: \: \: \: \: (\therefore \: adding \: ( \frac{1}{2} \angle \: B + \frac{1}{2} \angle \: D) \: on \: both \: sides) \\\end{gathered}⇒∠A+∠C+∠B+∠D+∠P+∠Q=360°+21∠B+21∠D(∴adding(21∠B+21∠D)onbothsides)</p><p></p><p>\begin{gathered}⇒ 360\degree + \angle \: p + \angle \: Q = 360\degree + \frac{1}{2} (\angle \: B + \angle \: D) \\ \: \: \: \: \: \: \: \: (\therefore \: sum \: of \: all \: the \: angles \: of \: a \: quadrilateral \: is \: 360\degree)\end{gathered}⇒360°+∠p+∠Q=360°+21(∠B+∠D)(∴sumofalltheanglesofaquadrilateralis360°)</p><p></p><p>\begin{gathered}⇒ \angle \: P + \angle \: Q = \frac{1}{2} (\angle \: B + \angle \: D). \\\end{gathered}⇒∠P+∠Q=21(∠B+∠D).</p><p></p><p>$

Hope it helps ❣️