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 of at point A and B and touches the circle with Centre p and. Prove angle ADE + angle BCEis equal to 180 degree​

Answer 1

Proved that, ∠ADE + ∠BCE = 180°.

Question :

In the figure two circles with Centre O and P intersect each other at point B and C . Chord AB of circle with centre O touches the circle with centre P in point E. Prove angle ADE + angle BCE is equal to 180 degree.                                    

To prove : ∠ADE + ∠BCE = 180°

Given :

"O" and "P" is the centre of the two circles.  

"O" and "P" intersects each other at point "B" and "C".

From the figure,

Note : Figure is attached below.

First join CD.

∠CEB and ∠CDE are in the alternate segment.

∴ ∠CEB = ∠CDE   ------> ( 1 )

Here,

• ABCD is a cyclic-quadrilateral.
• In cyclic-quadrilateral opposite sides are supplementary.

In ABCD cyclic-quadrilateral,

∠ADC + ∠ABC = 180°   -----> ( 2 )

Linear pair

∠CBE + ∠ABC = 180°    ------> ( 3 )

From equation ( 2 ) and ( 3 ), we get

∠ADC + ∠ABC = 180°   -----> ( 2 )

∠CBE + ∠ABC = 180°   ------> ( 3 )  

∠ADC = ∠CBE  ------> ( 4 )

In ΔCBE,

By angle sum property,

∠CBE + ∠CEB + ∠BCE = 180°

Using equation ( 1 ) and ( 2 ), we get

∠ADC + ∠CDE + ∠BCE = 180°

(∠ADC + ∠CDE) + ∠BCE = 180°

∠ADE + ∠BCE = 180°

Hence, proved that ∠ADE + ∠BCE = 180°.

To learn more...

1. 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

brainly.in/question/8363769

2. Two circles with centres O and O' intersect at points A and B. A line PQ is drawn to O O' through A (or B) intersecting the circles at P and Q. Prove that PQ = 2OO'

brainly.in/question/6592332