Circle geo Q. (1 Viewer)

[Saintly Devil]

Hi,

I couldn't figure this question out.

PQ, RS and LM are three parallel chords in a circle. Prove that triangle PRL is congruent to triangle QSM.

Could someone show me how to do it?

Note: Circle geometry is my worst topic - this is probably an easy question.

Thanks.

 

[BlackJack]

Long proof, 'cause I probably forgot half the shortcuts but:
(I assume respectivity in question: i.e. chord RS is between the other two.)

In cyclic quad PQSR:
>QPR = 180 - >QSR (PQSR cyclic)
>QSR = 180- >PQS (coint >'s, PQ || RS)
T: (Therefore) >QPR = >PQS
Similarly, in cyclic quad PQML with QP || ML:
>QPL = >PQM *
T: >QPR - >QPL = >PQS - >PQM
But >QPR - >QPL = >LPR
and >PQS - >PQM = >SQM
T: >LPR = >SQM (1)

Note: The same can be used for >SMQ = >PLR

Consider triangle QLP & QMP:
PQ common
>QPL = >PQM ( see*)
>QMP = >QLP (>'s subt. by chord QP in same segmt =)
T: QMP == QLP (AAS)
T: QM = PL (3)

T: since In triaingles. QSM and PRL:
(1), (2), (3) are true
QSM == PRL (AAS)

ED: spelling / turn HTML off next time...

 

[Saintly Devil]

Thanks for that......

I had been trying to figure it out for quite a while.
I totally suck at circle geometry.

 

[spice girl]

Here's a nice little alternative proof:

Consider the centre of circle, O, and consider the line, l, through O, perpendicular to all 3 chords PQ, RS, LM, and intersects these chords at A, B, C respectively.

Since l is perpendicular to all 3 chords, and l passes thru centre of circle, l bisects all three chords.

Thus PA = AQ, RB = BS, LC = CM

So the reflection of points P, R, L, about line l, become Q, S, M respectively (since PQ, RS, LM are perpendicular to line l)

Therefore triangle QSM is a reflection of triangle PRL and is thus congruent to triangle PRL.

Is this adequate?

 

[Dumbarse]

sounds pretty kool
never heard of this reflection method before but
does it count as a proof?

 

[BlackJack]

That was good, I was looking at it that way once, but I didn't think it was adequate...

But this probably is enough...

 

[spice girl]

Originally posted by Dumbarse
does it count as a proof?

I've NFI. It would work for 4u, but 3u? I've to ask my teacher.

 

[-=«MÄLÅÇhïtÊ»=-]

my method was abit like urs spicegal

i showed that the 3 parallel lines made 2 regular trapeziums (by using coint. angles and cyclic quad to show dat base angles equal)
and then i said that opposite sides of regular trapeziums are equal. Finally the far opposite angles are equal (since base angles are equal, 2 pairs of base angles also equal)

But its kinda unorthodox...

Like can u actually say that there are 2 parallel lines (eg. PQ and RS) and 2 sides (eg. PR and QS) are inclined at the same angle of incidenct crossing the parallel lines (<PRS and <QSR), so therefore PR and QS are equal?

 

[-=«MÄLÅÇhïtÊ»=-]

Like can u actually say that there are 2 parallel lines (eg. PQ and RS) and 2 sides (eg. PR and QS) are inclined at the same angle of incidence crossing the parallel lines (<PRS and <QSR), so therefore PR and QS are equal?

 

[-=«MÄLÅÇhïtÊ»=-]

gawd! i dunno why dat msg isn't coming out

its meant to read
continuation:
..parallel lines (PQ and RS), so there they (PR and QS) are equal

 

[BlackJack]

The HTML stuff interprets 'smaller than' signs as open tags and then since it doesn't work, ignores the word... that's my theory, 'coiz I had to went through my poof again changing them all to ">"

 

[Lazarus]

Sorry about that, I don't think there's any way around it short of disabling HTML codes in this forum.

 

[underthesun]

is this kind of question going to be in the 3u paper?

most these kinds will appear at 4u, but i hope it won't for 3u