Question

In the figure , two circles with centre o and p intersect each other at points the and c line intersect the circle with centre o at points a and b and touches the circle with centre p at point p prove that angle ade + angle Bce equals to 180 degree

Answer 1

Given the two circles with center o and p intersect each other at points and c line intersect the circle with center o at points a and b and touch the circle with center o at points a and b and touches the circle with center p at point p, angle ade + angle bce equals to 180 degrees.

• Given,
• The centers, O and P intersect each other at points C and D.
• Construction,
• Join C and D.
• To prove,
• ∠ADE+∠BCE=180°
• Proof,
• ∠CBE+∠ABC=180°      (linear pair)
• ∠ADC+∠ABC=180° (opposite angles of cyclic quadrilateral)
• from above equations, we get,
• ∠CBE=∠ADC
• and
• ∠CEB=∠CDE (angles in alternate segment)
• Now,
• In ΔCBE,
• ∠CBE+∠CEB+∠BCE=180° (sum of angles)
• ⇒[∠ADC+∠CDE]+∠BCE=180° (using the above equations)
• ∴ ∠ADE+∠BCE=180°
• Hence proved.

Answer 2

Proved that ∠ADE + ∠BCE = 180°.

Question : In the figure, two circles with center O and P intersect each other in points C and D. Chord AB of circle with center O touches the circle with center P in point E. Prove that angle ADE + angle BCE = 180°

To prove : ∠ADE + ∠BCE = 180°.

Given :

There are two circles.

"O" and  "P" are the centres of the two circles.

They intersect each other in points C and D.

Chord AB of circle with center O touches the circle with center P in point E.

Now, join CD.

In ΔCEB and ΔCDE angles are in the alternate segments.

∴ ∠CEB = ∠CDE  -----> ( 1 )

In cyclic-quadrilateral ABCD,

Cyclic-quadrilateral opposite angles are supplementary.

∠ADC + ∠ABC = 180° -----> ( 2 )

By linear pair

∠CBE + ∠ABC = 180° -----> ( 3 )

From the equation ( 2 ) and ( 3 ),

∠ADC + ∠ABC = 180° -----> ( 2 )

∠CBE + ∠ABC = 180° -----> ( 3 )

We get,

∠ADC = ∠CBE -----> ( 4 )

In ΔCBE,

Using angle sum property,

∠CBE + ∠CEB + ∠BCE = 180°

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

∠ADC + ∠CDE + ∠BCE = 180°

∠ADE + ∠BCE = 180°.               [ ∵ ∠ADC + ∠CDE = ∠ADE ]

Hence, ∠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