Vertices B and C of the triangle ABC lie on the circumference of a circle . AB and AC cut the circumference of the circle at X and Y respectively . Show that angle CBX +angle CYZ =180

If AB=AC , show that BC is parallel to XY .

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- Oct 30th 2009, 02:52 AMthereddevilsdeductive geometry (2)
Vertices B and C of the triangle ABC lie on the circumference of a circle . AB and AC cut the circumference of the circle at X and Y respectively . Show that angle CBX +angle CYZ =180

If AB=AC , show that BC is parallel to XY . - Oct 30th 2009, 09:13 AMEekNah90

let the centre of the circle be O

let angle COX(major) be 2x

angle CBX = 0.5 angle COX(major) = x

(angle of circumference is 1/2 of the angle at centre)

angle COX ( minor) + angle COX(major) = 360 (angel at a point)

thereofre angle COX (minor) = 360 -2x

angle CYX = 0.5 angle COX(minor) = 180 -x

(angle of circumference is 1/2 of the angle at centre)

so angle CBX + angle CYX = x + 180 -x = 180 (proven) - Oct 30th 2009, 09:20 AMSoroban
Hello, thereddevils!

The first part has. . .*theorem*

Quote:

Vertices and of lie on the circumference of a circle.

and cut the circumference of the circle at and respectively.

Show that: .

. . If a quadrilateral is inscribed in a circle, the opposite angles are supplementary.

Code:`X * * *`

o *

* *

* o Y

* *

* *

* o C

o *

B * *

* *

* * *

Draw chords

. .

Add [1] and [2]: .

. . . . . . . . . . . .

. . . . . . . . . . . .

- Oct 30th 2009, 11:12 AMukorov
(1) simply apply the theorem for inscribed quadrilateral....

or

(2) draw two radii XO and CO where O is the centre of circle. There is a theorem (forgotten its name) saying that__reflex__angle XOC = angle XYC x 2, and__obtuse__angle XOC = angle XBC x 2.

since the sum of the two angles XOC equals 360 degrees:

angle XYC x 2 + angle XBC x 2 = 360 degrees, hence

angle XYC + angle XBC = 180 degrees