help plzz (1 Viewer)

[polythenepam]

this question is driving me nuts
if anyone can do it that would be great
ABCD is a quadilateral such that <ABD=<DBC=<CDA=45
Q is a point on BD such that CQ bisects <BCA. Show that AQCD is a cyclic quad

 

[polythenepam]

ok my post got cut off for some reason
heres the full question
ABCD is a quadilateral such that angles ABD=DBC=CDA=45
Q is a point on BD such that CQ bisects angle BCA. show that AQCD is a cyclic quad.

 

[who_loves_maths]

Originally Posted by polythenepam
ABCD is a quadilateral such that angles ABD=DBC=CDA=45
Q is a point on BD such that CQ bisects angle BCA. show that AQCD is a cyclic quad.

hi polythenepam,

for your problem:
Let the diagonals BD and AC of quadralateral intersect at, say, the point T .
Now, consider triangle ABC, and label the following angles plz:
let < QCA = x = < BCQ, and < QAC = y, therefore < AQC = 180 -x -y (angle sum of triangle QAC)
< BTC = 180 -45 -2x (angle sum of triangle BCT), and so < BTA = 45 +2x (angle sum of straight line)
also, < CAB = 90 -2x = < CAQ + < QAB = y + < QAB ---> < QAB = 90 -2x -y
1) in triangles ATQ and CTQ, using the Sine Rule:
AQ/Sin(45+2x) = QT/Sin(y) ; and, CQ/Sin(180-45-2x) = QT/Sin(x) = CQ/Sin(45+2x)
divide these two equations to get ---> AQ/CQ = Sin(x)/Sin(y)
2) in triangles BAQ and BCQ, using the Sine Rule:
AQ/Sin(45) = BQ/< QAB = BQ/Sin(90 -2x -y) ; and, CQ/Sin(45) = BQ/Sin(x)
divide these two equations to get ---> AQ/CQ = Sin(x)/Sin(90 -2x -y)
hence, equating the above two equations for AQ/CQ:
AQ/CQ = Sin(x)/Sin(y) = Sin(x)/Sin(90-2x-y) ---> Sin(y) = Sin(90-2x-y) ---> y = 90-2x-y ---> y = 45 - x

therefore, < AQC = 180 -x -y = 135 ;
and so < AQC + < CDA = 135 + 45 = 180 ---> ie. ADCQ is a cyclic quadralateral (opposite angles are supplementary)

hope that helps

 

[polythenepam]

okaay thanx heeaaps..
i never thought of using trig rules.. thats a good idea.. do u think thats probably the only or shortest way to proving it.. cos i never would hav come up with that..
i kept trying to prove similarity but kept goin round in circles
ahh well
anyway thanx again

 

[shafqat]

who_loves_maths just outlined the method I was trying to use, but I mistakenly thought AB = BC, which made it too simple lol. i guess that's what studying for exams does to you.

 

[who_loves_maths]

Originally Posted by polythenepam
i never thought of using trig rules.. thats a good idea.. do u think thats probably the only or shortest way to proving it.. cos i never would hav come up with that..
i kept trying to prove similarity but kept goin round in circles
ahh well...

yes i think it's the shortest way if one would think of using trig at a moments notice. if you want to prove it using pure geometric methods such as using similarity, then you need to understand the basic underlying principle of similarity (eg. do you know how to PROVE the three similarity laws?)
to be able to prove them you actually need to introduce additional heights and lines in each of the triangles themselves... which is exactly what you need to do for this problem if you wish to use purely similarity to prove it. you are going around in circles because you haven't made any appropriate extra constructions.

but the thing is, thinking of drawing extra lines and forming extra triangles so you can use similarity takes as much time as thinking up of using trigonometry (which we've being using in geometry since year 8 or 9). so i think trig still provides a more direct method of tackling this particular problem.

 

[ngai]

ok my post got cut off for some reason
heres the full question
ABCD is a quadilateral such that angles ABD=DBC=CDA=45
Q is a point on BD such that CQ bisects angle BCA. show that AQCD is a cyclic quad.

In triangle ABC:
QC bisects angle C
QB bisects angle B
so Q is the incentre, and so QA bisects angle A

then u have:
BCQ=ACQ=x
ABQ=CBQ=45
BAQ=CAQ=45-x
so AQD = BAQ+ABQ = 45+45-x = 90-x
and CQD = CBQ+BCQ = 45+x
so AQC = 135-x+x = 135
also given ADC = 45
so AQCD cyclic (opp. angles supp.)

 

[who_loves_maths]

Originally Posted by ngai
In triangle ABC:
QC bisects angle C
QB bisects angle B
so Q is the incentre, and so QA bisects angle A

knowing that the three angle bisectors of a triangle is concurrent at its incentre is not part of the syllabus and is not taught in the course, even though the proof for it is not difficult.
i know you learn all about these things in special training for the elite, but spare a thought for the rest of us...

Edit: solving this problem using trig. is the equivalent of proving that the angle bisectors of a triangle are concurrent anyway.