Question

The bisectors of the angles of an acute-angled triangle ABC meet BC, CA, and AB at X, Y, and Z, respectively. Then

(A) BX ⋅ CY ⋅ AZ = XC ⋅ YA ⋅ ZB
(B) BX ⋅ AY ⋅ AZ = XC ⋅ CY ⋅ ZB
(C) BX ⋅ ZB ⋅ AZ = XC ⋅ YA ⋅ CY
(D) None of these

Answer 1

Option (A) is correct: BX * CY * AZ = XC * YA * ZB

Step-by-step explanation:

Let's prove option (A):

1. We can use the angle bisector theorem to find the ratios of the segments. Let BX/XC = b, CY/YA = c, and AZ/ZB = a.
2. Since the angle bisectors intersect at the incenter of the triangle, we know that the angles BXC, CYA, and AZB are all congruent.
3. Using the Law of Sines in triangles BXC, CYA, and AZB, we have:

• BX/AB = sin(BXC)/sin(ABC) = sin(CYA)/sin(BCA) = AZ/AB
• XC/AB = sin(XBC)/sin(ABC) = sin(YCA)/sin(BCA) = YA/AB

Combining these equations, we have BX/AB * XC/AB * AZ/AB = sin(BXC)/sin(ABC) * sin(XBC)/sin(ABC) * sin(YCA)/sin(BCA) = sin(BXC)*sin(XBC)*sin(YCA)/(sin(ABC)*sin(ABC)*sin(BCA))

Similarly, we can show that CY/BC * YA/BC * ZB/BC = sin(CYA)*sin(YCA)*sin(XBC)/(sin(BCA)*sin(BCA)*sin(ABC))

Since sin(BXC) = sin(XBC), sin(CYA) = sin(YCA), and sin(AZB) = sin(ZAB), we have BX/AB * XC/AB * AZ/AB = CY/BC * YA/BC * ZB/BC.

Therefore, option (A) is correct: BX * CY * AZ = XC * YA * ZB.