Question

Two circles with centre o and P intersect eachother in point C and D. Chord AB of the circlewith centre O touches the circle with centre Pin point E. (Fig. 11)Prove that ADE + BCE = 180Flg. 11​

Answer 1

Answer:

answe is 0 cauze we multiply0×1

Answer 2

Proved below.

Step-by-step explanation:

Given:

Here two circles having centre o and P intersect each other in point C and D. Chord AB of the circle having centre O touches the circle with centre P in point E.

Construction:

Join points C and D.

To prove:

∠ADE + ∠BCE = 180°

Proof:

Now, ∠CEB = ∠CDE       [1]   [angles in the alternate segment]

Now, ABCD is a cyclic quadrilateral, then∠ADC + ∠ABC = 180°  (Opposite angles of a cyclic quad. are supplementary)               [2]

Now, ∠CBE + ∠ABC = 180°  (Linear pair)                      [3]

From (2) and (3), we get

∠ADC = ∠CBE                        [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°

Hence proved.