Answer to Question #202078 in Calculus for Cynthia

Question #202078

in triangle ABC, the midpoints of the sides AB, BC and CA respectively. Show that 2AB + 3BC + AC =2LC


1
Expert's answer
2021-06-03T17:23:02-0400

To prove this, we construct the diagram below





With the lettering above, we have that

2AB+3BC+CA2\overrightarrow{AB} + 3\overrightarrow{BC} + \overrightarrow{CA} (1)

Note that

AB=2AL,  BC=BL+BC,  CA=CL+LA\overrightarrow{AB} = 2\overrightarrow{AL}, \space\space \overrightarrow{BC} = \overrightarrow{BL} + \overrightarrow{BC}, \space\space\overrightarrow{CA} = \overrightarrow{CL} + \overrightarrow{LA} (2)

Substituting (2) into (1), we have

2.2AL+3(BL+LC)+CL+LA2.2\overrightarrow{AL} + 3(\overrightarrow{BL} + \overrightarrow{LC} ) + \overrightarrow{CL} + \overrightarrow{LA}

This implies

4AL+3BL+3LC+CL+LA4\overrightarrow{AL} + 3\overrightarrow{BL} + 3\overrightarrow{LC} + \overrightarrow{CL} + \overrightarrow{LA} (3)

Again,

BL=AL,CL=LC,LA=AL\overrightarrow{BL} = − \overrightarrow{AL}, \overrightarrow{CL} = − \overrightarrow{LC} , \overrightarrow{LA}= − \overrightarrow{AL} (4)

Substituting equation (4) into equation (3), you will obtain that:

2AB+3BC+CA=4AL3AL+3LCLCAL2\overrightarrow{AB} + 3\overrightarrow{BC} + \overrightarrow{CA} = 4\overrightarrow{AL} − 3\overrightarrow{AL} + 3\overrightarrow{LC} − \overrightarrow{LC} − \overrightarrow{AL}

Collecting like terms, you obtain that;

2AB+3BC+CA=4AL3ALAL+3LCLC=LCLC=2LC2\overrightarrow{AB} + 3\overrightarrow{BC} + \overrightarrow{CA} = 4\overrightarrow{AL} − 3\overrightarrow{AL} − \overrightarrow{AL} + 3\overrightarrow{LC} − \overrightarrow{LC} = \overrightarrow{LC} −\overrightarrow{LC} = 2\overrightarrow{LC}


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Comments

Cynthia
09.06.21, 04:23

Thanks for solving the equations

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