Question

7.
In the figure, two circles with centres O and P
intersect each other at points D and C. Line AB
intersects the circle with centre o at points
A and B and touches the circle with centre Pat
point E. Prove ZADE + BCE = 180". (HOTS)​

Answer 1

Proved below.

Step-by-step explanation:

Given:

Here, two circles with centers O and P  intersect each other at points D and C.

Line AB  intersects the circle with center O at points  A and B and touches the circle with center P at  point E.

Join CD.

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°