Circle question (3 Viewers)

[bored of sc]

Let angle ACB = x

Then angle ACB = angle CBA = x (base angles in isosceles triangle ACB)

Also, angle AEB = angle ACB = x (angles subtending the same segment)

Furthermore, angle DAE = AEB = x (alternating angles on parallel lines, AD and BE)

But because ABCD is a cyclic quadrilateral, angle CDA = 180 - angle ABC = 180 - x.

Consequently, angle CDA + angle DAE = 180-x + x = 180.

Thus DC is parallel to AE.

Wow! That was an amazing proof. Well done.

 

[JSBboag]

lol, oh boy

whats wrong with that?

 

[Fortian09]

hmm i have another quesiton to add
it's pretty easy but i cant seem to prove it...

The figure shows a semicircle with diameter AD. If the chords AB=BC=CD, show that BC||AD
its kinda like a half hexagon

 

[alcalder]

Equal chords subtend equal angles at the circumference of a circle.

Thus, angle BDA = angle BCA = angle CBD = angle CAD = angle BAC = angle BDC = x

Angles on a diameter are 90 deg. Thus angle ABD = angle ACD = 90 deg

In triangle ACD:

3x + 90 = 180

Co-interior angles between lines BC and AD:

angle BCD + angle CDA = 3x + 90 = 180
angle CBA + angle BAD = 3x + 90 = 180

Since co-interior angles between lines BC and AD are supplementary, then BC ll AD

 

[Fortian09]

hmm I have a problem with this question


In the figure, BA and DC are produced to meet at X.
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a) Prove that Triangle XAC~ Triangle XDB.
b) Hence show that XA.XB = XC.XD
If XA = 4cm, AB = 5cm and CD = 9cm, find XC.

 

[Aerath]

In the figure, BA and DC are produced to meet at X.

a) Prove that Triangle XAC~ Triangle XDB.
b) Hence show that XA.XB = XC.XD
If XA = 4cm, AB = 5cm and CD = 9cm, find XC.

a) In triangles XAC and XDB:
Angle X is common
Angle XAC = BDX (exterior angle of cyclic quad)
Angle XCA = XBD ( '' '' '' '' '' )
therefore, XAC and XDB are congurent (AAA)

b) Can't remember the exact name of the rule, but it's something like lines from exterior point...bleh, I can't remember it. But it's a circle geometry rule

c) Using the given formula
XA*XB = XC*XD
4*9 = x*(x+9)
x^2 + 9x = 36
x^2 + 9x - 36 = 0
(x-3)(x+12) =0 (x>0)
x = 3

 

[shinn]

for b) i think they are asking you to prove that 'rule' (product of intercepts of intersecting secants)

So since triangle XAB /// triangle XDB,

=> XA / XD = XC / XB (ratio of sides equal in similar triangles)

Therefore, XA XB = XC XD

 

[Fortian09]

More questions...

 

[bored of sc]

In the parabola, the coordinates of C are (0, 1) and the coordinates of the vertex (V) are (2, -3).​

I hope...

Working: for coordinates of C
C is y-intercept therefore y-value (constant term from equation) is 1 and x-value is 0, thus (0, 1).

Working: for coordinates of vertex
From equation: a = 1, b = -4, c = 1.
x value(axis of symmetry) = -b/2a = --4/(2x1) = 4/2 = 2
Y-value: sub x = 2 into equation
y = (2)² -4(2) + 1 = 4 - 8 + 1 = -3
Therefore coordinates of the vertex are (2, -3).

 

[Aerath]

Minor segment: 0.5*r²(a-sina)
a = alpha



For parabola y =x²-4x+1
Vertex: x = -b/2a = 4/2 = 2
y = 2² - 4(2) + 1
= 4 - 8 + 1
= -3

Therefore vertex (2,-3)

A and B are x intercepts: xint when y = 0
x²-4x+1=0

Delta = 16-4(1)(1) = 12
x = (4+- 2r3)/2
= 2+-r3
Therefore A (2-r3, 0)
B (2+r3, 0)

C is y intercept, yint when x = 0
y = 1
C (0,1)

 

[lyounamu]

More questions...

1. Minor segment ACB = 1/2 r^2 (@ - sin@)

where r = radius and @ = angle in radian

2. c is when x=0, therefore y = 0^2 - 4 . 0 + 1 = 1
Therefore, c = (0,1)

V is the minimum point and it occurs when dy/dx = 0
i.e. dy/dx = 2x - 4 = 0
x = 2
Since x=2, y = 2^2 - 4.2 + 1 = -3

V = (2,-3)

3. First prove that ABC ll ADE (equiangular)

Then, divide ABC by ADE. So you have 9 pieces. 4 pieces make up the BDEF. Therefore, the ratio is
1:4

 

[Fortian09]

what about the triangle one?

 

[lyounamu]

what about the triangle one?

I did it. look at my edited one.

 

[Fortian09]

ahh tyty Iyounamu

 

[lsam]

1. Minor segment ACB = 1/2 r^2 (@ - sin@)

where r = radius and @ = angle in radian

2. c is when x=0, therefore y = 0^2 - 4 . 0 + 1 = 1
Therefore, c = (0,1)

V is the minimum point and it occurs when dy/dx = 0
i.e. dy/dx = 2x - 4 = 0
x = 2
Since x=2, y = 2^2 - 4.2 + 1 = -3

V = (2,-3)

3. First prove that ABC ll ADE (equiangular)

Then, divide ABC by ADE. So you have 9 pieces. 4 pieces make up the BDEF. Therefore, the ratio is
1:4

omfg, namu.

don't you get sick of doing all those maths questions seriously?

80% of the posts related to maths included you.

 

[lyounamu]

omfg, namu.

don't you get sick of doing all those maths questions seriously?

80% of the posts related to maths included you.

Well. Doesn't this indicate that I am working hard enough to be your worthy opponent? Better get pumped up, mate. Or answer some questions here, help me out. Lol

 

[tommykins]

omfg, namu.

don't you get sick of doing all those maths questions seriously?

80% of the posts related to maths included you.

Sorry if this is off topic - but for your sig (HSC 2007)

does that mean you achieved those marks in year 10?

 

[lyounamu]

Sorry if this is off topic - but for your sig (HSC 2007)

does that mean you achieved those marks in year 10?

Yeah, Sam did his two hsc exams last year. (lucky bastard)

 

[Fortian09]

hey how does that dividing triangle thing work?

 

[lolokay]

hey how does that dividing triangle thing work?

ABC is 3^2=9 times bigger than ADE, since they are similar and AC is three times bigger than AE. Similarly EFC is 4 times bigger than ADE, so BDEF is 9-4-1=4 times bigger.