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 .
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 . . .
Theorem: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
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Add [1] and [2]: .
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(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