# part one of the hard question trigiolgy

• Jun 12th 2007, 11:57 PM
calculus_jy
part one of the hard question trigiolgy
P, Q are points on the side AC, AB of triangle ABC such that
angBPC = ang BQC; BP cuts CQ at K; X, Y are points on AC, AB such that KXAY is a parallelogram. Prove that (i) AX.XC = AY.YB (ii) the centre of the circle ABC is equidistant from X and Y

I hav already solve (i) but just to typed if it is of any help to (ii)
PS sorry but i dont know how to draw diagrams on computer
• Jun 13th 2007, 03:24 AM
OReilly
Quote:

Originally Posted by calculus_jy
P, Q are points on the side AC, AB of triangle ABC such that
angBPC = ang BQC; BP cuts CQ at K; X, Y are points on AC, AB such that KXAY is a parallelogram. Prove that (i) AX.XC = AY.YB (ii) the centre of the circle ABC is equidistant from X and Y

I hav already solve (i) but just to typed if it is of any help to (ii)
PS sorry but i dont know how to draw diagrams on computer

What does this means AX.XC = AY.YB !?
• Jun 14th 2007, 02:09 AM
calculus_jy
Speification
AX.XC = AY.YB means the length of AX multiple by the length of XC equals the length of AY multiple by the length of YB
• Jun 14th 2007, 05:37 AM
OReilly
Quote:

Originally Posted by calculus_jy
AX.XC = AY.YB means the length of AX multiple by the length of XC equals the length of AY multiple by the length of YB

Can you post that proof?

• Jun 14th 2007, 11:11 PM
calculus_jy
proof
since angBQC = angCPB (given) and are on the same side of BC, the pts BQPC are concylic (equal angles subtended on same side of a line)

Therefore, angYBK=angQBK = ang XCK=angXCK (angles subtended on same arc QP)

In parallelogram AYKX, angAYK = angAXK (opp ang of parallelgram)
Therefore angBYK = 180 deg - angAYK = 180 deg - angAXK = angCXK

Then it can be proven with angBYK = angCXK and angYBK = angXCK that the triangles YKB and XKC are equiangular

thus YK / XK = YB / XC (sides proportional in eqiangular tri)

YK . XC = YB . XK

But YK = AX (opp sides of parallelogram) and similiarly AY= XK

AX. XC = AY. YB

Hope that helps to do part (ii) because i really hav no idea
• Jun 15th 2007, 09:37 AM
OReilly
Quote:

Originally Posted by calculus_jy
since angBQC = angCKP and are on the same side of BC, the pts BQPC are concylic (equal angles subtended on same side of a line)

Angles BQC and CKP are not equal, I don't know how did you get that?
• Jun 16th 2007, 10:38 PM
earboth
Quote:

Originally Posted by calculus_jy
P, Q are points on the side AC, AB of triangle ABC such that
angBPC = ang BQC; BP cuts CQ at K; X, Y are points on AC, AB such that KXAY is a parallelogram. Prove that (i) AX.XC = AY.YB (ii) the centre of the circle ABC is equidistant from X and Y
I hav already solve (i) but just to typed if it is of any help to (ii)
PS sorry but i dont know how to draw diagrams on computer

Hello,

I've attached a sketch of this problem.

1. The angles $\angle(BQC)$ and $\angle(BPC)$ are equal and they are subtended over the same side of the triangle thus the points B, Q, P, C must lie on a circle with $M_2$ as centre (green).

2. Therefore you can use the intersecting chord theorem here, using BP and CQ as chords and K as intersection.

3. The triangles YQK and PXK are similar.

4. With the green circle and the secants AC and AB you can use the secant-secant-theorem.

5. The two centres $M_1$ and $M_2$ must lie on a perpendicular bisector of BC. And $M_1$ must lie on a perpendicular bisector of XY.

But: I haven't got a clue how to put all those facts together to proof the statement of your problem.
• Jun 18th 2007, 12:15 AM
calculus_jy
Thanks
regarding to oreilly, it was a typing mistake that i made.

earthboth, i dont get what that intersecting chord theorem in 2 is used for, also i dont get what the secant-secant theorm is and used for either,
however, it will be nice if u can provide the proof that
http://www.mathhelpforum.com/math-he...baaa6990-1.gif must lie on a perpendicular bisector of XY as this is the prove that the centre is eqidistant from XY

Thank you so much for both ur help