Question
In $\triangle\text{ABC, }\angle\text{B}=35^{\circ},\angle\text{C}=65^{\circ}$ and the bisector of $\angle\text{BAC}$ meets $BC$ in $X.$ Arrange $AX, BX$ and $CX$ in descending order.
✓
Answer
Given: in $\triangle\text{ABC, }\angle\text{B}=35^{\circ},\angle\text{C}=65^{\circ}$ and the bisector of $\angle\text{BAC}$ meets $BC$ in $x$ In $\triangle\text{ABX},$
$\because\angle\text{BAX}>\angle\text{ABX}$
$\therefore\text{BX}>\text{AX}...(\text{i})$
$\therefore\text{AX}>\text{CX}...(\text{ii})$ From $(i)$ and $(ii),$ we get $\text{BX > AX > CX}$
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