In ∆ABC, B=35°,C=65° and the bisector of BAC meets BC in X. Arrange AX, BX and CX in descending order.

Given: In ∆ABC, B=35°,C=65° and BAX = XAC


To find: Relation between AX, BX and CX in descending order.


In ∆ABC, by the angle sum property, we have


A + B + C = 180°


A + 35° + 65° = 180°


A + 100° = 180°


A = 80°


But BAX = A


= × 80° = 40°


Now in ∆ABX,


B = 35°


BAX = 40


And BXA = 180° - 35° - 40°


= 105°


So, in ∆ABX,


B is smallest, so the side opposite is smallest, ie AX is smallest side.


AX < BX …(1)


Now consider ∆AXC,


CAX = × A


=× 80° = 40°


AXC = 180° - 40° - 65°


= 180° - 105° = 75°


Hence, in ∆AXC we have,


CAX = 40°, C = 65°, AXC =75°


∴∠CAX is smallest in AXC


So the side opposite to CAX is shortest


Ie CX is shortest


CX <AX …. (2)


From 1 and 2 ,


BX > AX > CX


This is required descending order


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