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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