a rather obstinate 3u Geometry question (not circle). (1 Viewer)

[who_loves_maths]

okay ppl, for those keen 3u (or 4u) mathematicians, plz attempt this question:

In triangle ABC, the angle bisector of < ABC from B meets the side AC at B'. and the angle bisector of < ACB from C meets the side AB at C'. If BB' = CC' , then prove that triangle ABC is isosceles.

this question is from the green yr12 Cambridge Mathematics Ext.1 book (under Chapter Euclidean Geometry).

the problem is that i did the question successfully... but it took two pages of rigorous algebra... which is what i'm not very happy about. (i didn't use co-ordinate geometry, and plz don't use it either.)

so if anyone out there already know a simpler method of doing this question, or can find one themselves, then plz post your solutions...

Thanks in advance, any help is greatly appreciated.

 

[acmilan]

er ill give it a shot tomorrow when i finish my final physics exam

 

[JamiL]

okay ppl, for those keen 3u (or 4u) mathematicians, plz attempt this question:

In triangle ABC, the angle bisector of < ABC from B meets the side AC at B'. and the angle bisector of < ACB from C meets the side AB at C'. If BB' = CC' , then prove that triangle ABC is isosceles.

this question is from the green yr12 Cambridge Mathematics Ext.1 book (under Chapter Euclidean Geometry).

the problem is that i did the question successfully... but it took two pages of rigorous algebra... which is what i'm not very happy about. (i didn't use co-ordinate geometry, and plz don't use it either.)

so if anyone out there already know a simpler method of doing this question, or can find one themselves, then plz post your solutions...

Thanks in advance, any help is greatly appreciated.

i dunno but as far as i figure, B'C'CB is a rombus, with B'C' parrall 2 BC, since the vertacies are equal. and therefor angle B'BC = angle BCC'... amking it a isosalies triangle. now i dunno if that is 100% correct or not

 

[who_loves_maths]

Originally Posted by JamiL
i dunno but as far as i figure, B'C'CB is a rombus, with B'C' parrall 2 BC, since the vertacies are equal. and therefor angle B'BC = angle BCC'... amking it a isosalies triangle. now i dunno if that is 100% correct or not

hmm... what do you mean B'C'CB is a rhombus? if it is then all sides must equal in length, and clearly B'C' doesn't equal BC.

and can you also explain what you mean by "...since the vertices are equal"?


EDIT: okay, i think i know what you mean, you got a good point here (i didn't think of that before). but the problem still remains to prove that B'C'BC is a TRAPEZIUM {which is probably just as hard as proving the angle are equal without using properties of the trapezium anyways, since both results are directly linked to each other.}

 

[Mill]

Turn it into a circle geometry question.

Construct a circle passing through BCB'C'.

Label angles C'BB' and CBB' as x (since they are equal as bisecetion is given).
Label angles BCC' and B'CC' as y (since they are equal as bisecetion is given).

Then it remains to prove that 2x = 2y (or x = y) in order to prove that ABC is isosceles.

Now in the circle, chord BB' subtends angle 2y at the circumference and chord CC' subtends angle 2x at the circumference. Since we know that equal chords subtend equal angles at the circumference, we can conclude 2x = 2y, and thus that the triangle ABC is isosceles.

 

[shafqat]

You've assumed BC'B'C is a cyclic quadrilateral.

 

[who_loves_maths]

Originally Posted by Mill
Turn it into a circle geometry question.

Construct a circle passing through BCB'C'.

Label angles C'BB' and CBB' as x (since they are equal as bisecetion is given).
Label angles BCC' and B'CC' as y (since they are equal as bisecetion is given).

Then it remains to prove that 2x = 2y (or x = y) in order to prove that ABC is isosceles.

Now in the circle, chord BB' subtends angle 2y at the circumference and chord CC' subtends angle 2x at the circumference. Since we know that equal chords subtend equal angles at the circumference, we can conclude 2x = 2y, and thus that the triangle ABC is isosceles.

lol, you think i haven't tried that approach yet? in fact, circle geometry was the first thing i tried with that question... but i still ended up solving it using Euclidean geometry of the triangle anyway.

i found that whatever i did with the circle geo stuff on the triangle, it alway has an equivalent in normal Euclidean geo anyways... and in the end, like you said, you're just trying to prove that x =y, which can be done without the need of the circle.

also, like shafqat just commented, you can't just assume BC'B'C is a cyclic quad! and to prove it is, you need to prove x =y, so it's a catch-22.

 

[Mill]

Very true :0 Amateur mistake! Disregard what I said