deductive geometry (2)

thereddevils · #1 $Old$ October 30th, 2009, 04:52 AM

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 .

EekNah90 · #2 $Old$ October 30th, 2009, 11:13 AM

Quote:

Originally Posted by thereddevils $View Post$

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)

Soroban · #3 $Old$ October 30th, 2009, 11:20 AM

Hello, thereddevils!

The first part has theorem . . .

Quote:

Vertices $B$ and $C$ of $\Delta ABC$ lie on the circumference of a circle.

$AB$ and $AC$ cut the circumference of the circle at $X$ and $Y$ respectively.

Show that: . $\angle CBX + \angle CYX \:=\:180^o$

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 $XY,\: YC,\: CB,\: BX.$

. . $\begin{array}{cccccccc}\angle Y \:=\:\frac{1}{2}\text{arc}(XBC) & [1] \\ \\[-4mm] \angle B \:=\:\frac{1}{2}\text{arc}(XYC) & [2] \end{array}$

Add [1] and [2]: . $\angle Y + \angle B \;=\;\tfrac{1}{2}\text{arc}(XBC) + \tfrac{1}{2}\text{arc}(XYC)$

. . . . . . . . . . . . $\angle Y + \angle B \;=\; \tfrac{1}{2}\bigg[\text{arc}(XBC) + \text{arc}(XYC)\bigg] \;=\;\tfrac{1}{2}(360^o)$

. . . . . . . . . . . . $\angle Y + \angle B \;=\;180^o$

ukorov · #4 $Old$ October 30th, 2009, 01:12 PM

(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