Questions [Circle Geometry] (1 Viewer)

[FriedRice07]

Okay so I encountered more problems while doing these questions so feel free to lend a hand

Question 1)
ABCD is a cyclic quadrilateral in which the opposite sides AB and DC are equal.

a) Draw a diagram. (It's a square in a circle right?)
b) Prove that the diagonals AC and BD are equal.

Question 2a)

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AB is a chord of a circle and CAD is a tangent to the circle at the point A. The bisector of angle BAC meets the circle again at P and the bisector of angle BAD meets the circle again at Q.

Show That:
a) PQ is a diameter of the circle.
b) PQ is perpendicular to the chord AB.

Question 2b) (applies to above diagram)
PQ meets AB at R and BQ produced meets CD at S. If BS is perpendicular to CD, prove that:

a)

BAD = 60°.
b) QR = QS.
c) AB = AP.

Question 3)
In the diagram below, AB is a diameter of a circle, whose center is the point O. The chord XY passes through M, the mid-point of OB. AX and BY are joined.

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a) Prove the two triangles formed (triangles AXM and MYB) are similar.
b) If XM = 8 cm and YM = 6 cm, find the length of the radius.

Question 4)
ABC is a triangle inscribed in a circle. The tangent at A meets BC produced at D.

DAC = 40°.

CDA = 10°

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a) Show that BC is a diameter of the circle.
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[white ferret]

1a) Not necessarily. it can be a trapezium. I think AD and BC should be parallel. Don't quite rmb the rules about the properties of the shapes.. Dig those up, and u should be able to figure it out.

2a) Let PAC (dot) = x
RAQ (cross) = y
2x + 2y = 180 (straight angle)
x + y = 90
RAP + BAQ = 90
PQ is diameter (angle in a semicircle = 90)

3a) AMX = YMB (vert opp)
AXY = ABY (angles standing on same arc)
XAB = XYB (" ")
similar because they are equiangular.

b) Let MB = x
AM = 3x
(3x)/6 = 8/x
3x^2 = 48
x = 4
Diameter = 4x = 16cm

4. DCA = 130 (angle sum triangle)
ACB = 50 (suppl angles)
ABC = 40 (angle in alt seg)
BAC = 90 (angle sum triangle)
BC is diameter because angle is a semicircle is 90

 

[FriedRice07]

4. DCA = 130 (angle sum triangle)
ACB = 50 (suppl angles)
ABC = 40 (angle in alt seg)
BAC = 90 (angle sum triangle)
BC is diameter because angle is a semicircle is 90

Thanks

One down, three to go !

 

[Aerath]

Didn't white ferret do all the questions? =\

Edit: Saw the edit time.

 

[lolokay]

2a) b) Draw in perpendicular to tangent at A to a point E, through origin O. This is the diameter of the circle.

OAP = x
OPA = x (isos triangle)
POA = 180 - 2x
ROA = 2x
OAR = 90 - 2x
.: PRA = 90, so the lines are perpendicular

2b)a)
RQB = o = 90 - x
2x + 90 + 90 + 180 - (90 - x) = 360
3x = 90
2x = 60 = BAD

b) triangles ARQ and ASQ are equiangular with AQ common, so congruent
therefore QR = QS

c) o = 90 - x = 60
[AE is tangent at A]
in triangles APE and ABE
AE is common
EAB = 90 - 2x = 30
EAP = 90 - o = 30
EAB = EAP
APE = ABE = 90
therfore congruent,
so AB = AP