Question

the circle which passes through vertices A and C of parellogram OABC has centre O .if BA produced meets the circle at D ,then prove that

Answer 1

CORRECT QUESTION :

the circle which passes through the vertices A and C of a parallelogram OABC has center at O. if BA produced meets the circle at D, prove that angle ABC= 2 times angle OCD

REQUIRED ANSWER :

$\implies$ABCD is a parallelogram.

$\implies$Opposite angles are same.

$\implies$∠ABC = ∠ADC

$\implies$∠BAD = ∠BCD

$\mapsto$ A circle is drawn in such a way that it passes through A, B and C.

$\mapsto$ P is a point on circle intersecting with line CD.

$\mapsto$ ABCP is cyclic quadrilateral.

(Circle is passing through all four vertices. A, B, C and P).

$\mapsto$ From the property of Cyclic Quadrilateral, ∠ABC + ∠APC = 180°.

(Opposite angles are supplementary)

$\mapsto$ ∠APC + ∠APD = 180°. (Angle on straight line)

From above 2, we can derive ∠ABC = ∠APD.

$\mapsto$ ∠ABC = ∠ADP = ∠APD.

$\mapsto$ ∠ADP = ∠APD.

Answer 2

$\huge\mathcal\colorbox{lavender}{{\color{b}{✿Yøur-Añswer♡}}}$

$\large\bf{\underline{\red{GIVEN}}}$

the circle which passes through vertices A and C of parellogram OABC has centre O .if BA produced meets the circle at D

$\large\bf{\underline{\red{TO\:FIND}}}$

∠ADP = ∠APD.

$\large\bf{\underline{\red{SOLUTION}}}$

Question is to prove that ∠APD = ∠ADP

See the attached picture to represent the problem.  

ABCD is a parallelogram.  

=>Opposite angles are same.  

=>∠ABC = ∠ADC

=>∠BAD = ∠BCD

A circle is drawn in such a way that it passes through A, B and C.  

P is a point on circle intersecting with line CD.

ABCP is cyclic quadrilateral.  

(Circle is passing through all four vertices. A, B, C and P).

From the property of Cyclic Quadrilateral, ∠ABC + ∠APC = 180°.

(Opposite angles are supplementary)

∠APC + ∠APD = 180°. (Angle on straight line)

From above 2, we can derive  ∠ABC = ∠APD.  

∠ABC = ∠ADP = ∠APD.  

∠ADP = ∠APD.

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