Circles =] (1 Viewer)

[ttong]

Hey all,
im having trouble with the questions i attached, specifically the last two parts of both qs. pls help

thx in advance

 

[lyounamu]

Hey all,
im having trouble with the questions i attached, specifically the last two parts of both qs. pls help

thx in advance

For ii)

Construction: join BC, BR, AC and AP.

AB is a straight line because C is a point of intersection and the line joining the circles' centres pass through their point of contact which is C.

Then we have two isoceles triangles. So they are similar because the ratio will be similar as well.

Now, it can be proved that angle BCR = angle ACP (corresponding angles are similar)

Hence PCR is a straight line because angle BCR = angle ACP proving the theorem correct (vertically opposite angles are equal).

iii)

Construction: join BD and CP.
By joining BD you created two equal triangles BCD and BQD because BC = BQ and BD is a common side and angle CBD = angle BQD (SAS)

However for now, let's just look at triangle BCR. Let BCR = x. So CBR is 180-2x then CBQ = 2x. Since we proved that angle CBD = angle BQD, angle CBD = x.

Then angle BDQ = 90 - x

However, CPD is actually 90-x because APD = 90 (angle between a radius and tangent = 90)
So angle CPD = angle BDQ. So BD and CP are parallel.

iv)
Angle TSP = 90 (angle in a semicircle = 90)
So angle PSR = 90
since angle PSR + angle RQP = 180, the SRQP is a cyclic (?) quadrilateral which means that they are concyclic in an imaginary circle.

 

[bored of sc]

Nice work Namu. :wave:

 

[lyounamu]

Nice work Namu. :wave:

Haven't finished yet. It's the longest circle question that I have ever seen. (4 bloody parts!)

2nd Q)

i)
angle BFH = 90.
angle HDB = 90.

Opposite angles add to 180 so it BFHD is a cyclic quadrilateral.

Cannot prove te AFDC one. My brain is not functioning well. Cannot even darw an accurate diagram.
Shall move on to ii)

Ahhh~ damn it. I cannot handle both the maths and english assignment simultaneously and my brain is not working well as I mentioned. Somebody do the 2nd question for me. (I mean for the OP). I am off to english. (sorry maths, eng is more important today)

 

[lolokay]

for the second one, would we be allowed to use the rule that the altitudes of a triangle intersect at a common point?

 

[lyounamu]

for the second one, would we be allowed to use the rule that the altitudes of a triangle intersect at a common point?

I am not really sure. I thought about that but I really don't know whether I can use that to this particular question. I will see other alternatives if I can think of.

 

[lolokay]

I suppose if there's nothing else you can do, you could first prove that the altitudes are concurrent, as shown here - http://jwilson.coe.uga.edu/EMT668/E...nout/write-up 3.aarnout/orthocenterproof.html

 

[ttong]

thanks heaps lyounamu. btw u dont need to do that altitude thing for that q. jst found out how to do it and yeh theres an easier way.

 

[lyounamu]

thanks heaps lyounamu. btw u dont need to do that altitude thing for that q. jst found out how to do it and yeh theres an easier way.

That's awesome.

 

[lolokay]

how do you do it? (the show AFDC is cyclic bit)

 

[Mark576]

∠AFC = 90^o
∠ADC = 90^o

∴AFDC is cyclic [equal angles subtended on the same side of an interval]

 

[lolokay]

oh wow. how did I not see that?