Question

In the figure, two circles with centres O and P
intersect each other at points D and C. Line AB
2 intersects the circle with centre 0 at points A and
B and touches the circle with centre P at point E.Prove (angle) ADE +(angle)BCE =180°.
(HOTS)
​

Answer 1

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°

Answer 2

Step-by-step explanation:

SOLUTION ✍️

_________________________

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°

hope this helps you!!