# Thread: part one of the hard question trigiolgy

1. ## 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

2. 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 !?

3. ## 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

4. 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?

I have uploaded diagram.

5. ## 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

6. 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?

7. 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 $\displaystyle \angle(BQC)$ and $\displaystyle \angle(BPC)$ 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 $\displaystyle M_2$ 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 $\displaystyle M_1$ and $\displaystyle M_2$ must lie on a perpendicular bisector of BC. And $\displaystyle M_1$ 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.

8. ## 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
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