Question

in the figure two circles with Centre O and P intersect each other at point B and C then A B intersect the circle with Centre O at points A and B and touches the circle with Centre p at point P prove that angle A + angle b is equal to 180 degree ​

Answer 1

Answer:

In Truangle BCE

BCE+CBE+CEB=180

BUT, CBE=ADC-- EXTERIOR ANGLE OF CYCLIC QUADRILATERAL-

AND CEB= EDC ---2 (angles in tangent and secant)

From 1 and 2 we get

BCE+ADC+EDC=180

BCE+ADE=180. (angle additional property)

Hence proved

Answer 2

PROVED

Step-by-step explanation:

given :  two circles with Centre O and P intersect each other at point B and C

to prove : $\angle ADE + \angle BCE = 180^\circ$

proof : join CD

in triangle CEB and CDE angles are in the same segments

$\angle CEB = \angle CDE ...(1)$

we know that the in a cyclic quadrilateral ABCD

opposite angles are supplementary

thus ,

$\angle ADC + \angle ABC = 180^\circ$...(2)

by linear pair , $\angle CBE + \angle ABC = 180^\circ...(3)$

from 2 and 3

we get

$\angle ADC = \angle CBE ..(4)$

now , in triangle CBE

using angle sum property

$\angle CBE + \angle CEB + \angle BCE = 180^\circ \\\\\angle ADC + \angle CDE + \angle BCE = 180^\circ \\\\\angle ADE + \angle BCE = 180^\circ$

hence proved

#Learn more :

https://brainly.in/question/15935974