question_answer

In \[\Delta \]ABC, \[\angle \]B = \[35{}^\circ \], \[\angle \]C = \[65{}^\circ \] and the bisector of \[\angle \]BAC meets \[\overline{BC}\]in X. Arrange \[\overline{AX},\overline{BX}\]and \[\overline{AB}\]in descending order.

A)  \[\overline{AB}=\overline{BX}=\overline{CX}\]

B)  \[\overline{AB}<\overline{BX}<\overline{CX}\]

C)  \[\overline{BX}<\overline{AB}<\overline{CX}\]

D)  \[\overline{AB}>\overline{BX}>\overline{CX}\]

Correct Answer: D

Solution :

\[\angle B=35{}^\circ ,\angle C=65{}^\circ \] \[\angle A=180{}^\circ -35{}^\circ -65{}^\circ =80{}^\circ \] \[\therefore \] \[\angle BAX=40{}^\circ =\angle XAC\] In \[\Delta \,ABX\] \[\angle B=35{}^\circ ,\angle BAX=40{}^\circ \] \[\angle BXA=105{}^\circ \] \[\therefore \] \[\overline{AB}>\overline{BX}>\overline{CX}\]