Question

32. The bisectors of the angles of an acute angled triangle ABC meets 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

Give Along With Solutions

Answer 1

Answer:

option (A) is correct

Step-by-step explanation:

$\textbf{Angle bisector theorem:}$

$\text{ When vertical angle of a triangle is bisected, }$

$\text{the bisector divides the base into two segments }$

$\text{which have the ratio as the order of other two sides.}$

$\text{In }\triangle\,ABC,\text{AX is the bisectore of }\angle{A}$

$\text{By angle bisector theorem }$

$\frac{BX}{XC}=\frac{AB}{AC}$....(1)

$\text{In }\triangle\,ABC,\text{BY is the bisectore of }\angle{B}$

$\text{By angle bisector theorem }$

$\frac{CY}{YA}=\frac{BC}{AB}$....(2)

$\text{In }\triangle\,ABC,\text{CZ is the bisectore of }\angle{C}$

$\text{By angle bisector theorem }$

$\frac{AZ}{ZB}=\frac{AC}{AB}$....(3)

$\text{Multiplying (1)(2) and (3)}$

$\frac{BX}{XC}.\frac{CY}{YA}.\frac{AZ}{ZB}=\frac{AB}{AC}.\frac{BC}{AB}.\frac{AC}{AB}$

$\implies\,\frac{BX}{XC}.\frac{CY}{YA}.\frac{AZ}{ZB}=1$

$\implies\,\boxed{\bf\,AZ.BX.CY=AY.BZ.CX}$