To prove this, we construct the diagram below
With the lettering above, we have that
2AB+3BC+CA (1)
Note that
AB=2AL, BC=BL+BC, CA=CL+LA (2)
Substituting (2) into (1), we have
2.2AL+3(BL+LC)+CL+LA
This implies
4AL+3BL+3LC+CL+LA (3)
Again,
BL=−AL,CL=−LC,LA=−AL (4)
Substituting equation (4) into equation (3), you will obtain that:
2AB+3BC+CA=4AL−3AL+3LC−LC−AL
Collecting like terms, you obtain that;
2AB+3BC+CA=4AL−3AL−AL+3LC−LC=LC−LC=2LC
Comments
Thanks for solving the equations
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