Question

in the figure two circles with Centre O and P intersect each other at point B and C line a b intersect the circle with Centre O at point A and B and touches the circle with Centre p at point P prove that angle abc + Angle B C is equal to 180

Answer 1

ABCD is a cyclic quadrilateral where ∠ADE + ∠BCE = 180°

Construction - Join CD.

∠CEB =∠CDE  ....(1)    [As angles are in the alternate segment]

Since, ABCD is a cyclic quadrilateral, then

∠ADC + ∠ABC = 180°  (The opposite angles of cyclic quadilateral are supplementary)  --- eq 2

∠CBE + ∠ABC = 180°  (Linear pair)  --- eq 3

From equation (2) and (3), -

∠ADC = ∠CBE  --- eq 4

In ΔCBE,  

 ∠CBE+∠CEB+∠BCE = 180°  [Angle sum property]

=∠ADC + ∠CDE + ∠BCE = 180°  [Using 1 and 4 ]

= ∠ADC + ∠CDE]+ ∠BCE = 180°

= ∠ADE + ∠BCE = 180°

Answer 2

Answer:

Step-by-step explanation:

∠CEB =∠CDE  ....(1)    [As angles are in the alternate segment]

Since, ABCD is a cyclic quadrilateral, then

∠ADC + ∠ABC = 180°  (The opposite angles of cyclic quadilateral are supplementary)  --- eq 2

∠CBE + ∠ABC = 180°  (Linear pair)  --- eq 3

From equation (2) and (3), -

∠ADC = ∠CBE  --- eq 4

In ΔCBE,  

 ∠CBE+∠CEB+∠BCE = 180°  [Angle sum property]

=∠ADC + ∠CDE + ∠BCE = 180°  [Using 1 and 4 ]

= ∠ADC + ∠CDE]+ ∠BCE = 180°

= ∠ADE + ∠BCE = 180°