Thread: deductive 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 .

Originally Posted by thereddevils

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 CYX =180

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

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)

Hello, thereddevils!

The first part has theorem . . .

Vertices and of lie on the circumference of a circle.

and cut the circumference of the circle at and respectively.

Show that: .

Theorem:
. . 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]: .

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

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

(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