Respuesta :

Given :

  • Let ABCD Is Cyclic Quadrilateral

To Prove :

[tex] \gray{ \frak{ \angle A + \angle C = 180° }}[/tex]

[tex] \: \: [/tex]

[tex] \gray{\frak{\angle B + \angle D = 180°}}[/tex]

[tex] \: \: [/tex]

Construction Join OB & OD

Proof :

[tex] \: \: [/tex]

[tex] \gray{ \frak {\angle BOD = 2 \angle BAD}}[/tex]

[tex] \: \: [/tex]

[tex]\gray{\frak{ \angle BAD = \frac{1}{2} \angle BOD}}[/tex]

[tex] \: \: [/tex]

Similarly :

[tex] \: \: [/tex]

[tex]\gray{\frak{\angle BCD = \frac{1}{2} \angle DOB }}[/tex]

[tex] \: \: [/tex]

[tex]\gray{\frak{ \angle BAD + \angle BCD = \frac{1}{2}\angle BOD + \frac{1}{2} \angle DOB }}[/tex]

[tex] \: \: [/tex]

[tex] \gray{ \frak { = \frac{1}{2} ( \angle BOD + \angle DOB )}}[/tex]

[tex] \: [/tex]

[tex] \gray {\frak{ = ( \frac{1}{2}) \times 360 \degree = 180 \degree }}[/tex]

[tex] \: \: [/tex]

[tex] \sf similary : \boxed{\color{skyblue}{ \frak{ \angle \: B + \angle D = 180°}}}[/tex]

[tex] \: \: [/tex]

Hope Helps ! :)

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