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

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

Note that

"\\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

"2.2\\overrightarrow{AL} + 3(\\overrightarrow{BL} + \\overrightarrow{LC} ) + \\overrightarrow{CL} + \\overrightarrow{LA}"

This implies

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

Again,

"\\overrightarrow{BL} = \u2212 \\overrightarrow{AL}, \\overrightarrow{CL} = \u2212 \\overrightarrow{LC} , \\overrightarrow{LA}= \u2212 \\overrightarrow{AL}" (4)

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

"2\\overrightarrow{AB} + 3\\overrightarrow{BC} + \\overrightarrow{CA} = 4\\overrightarrow{AL} \u2212 3\\overrightarrow{AL} + 3\\overrightarrow{LC} \u2212 \\overrightarrow{LC} \u2212 \\overrightarrow{AL}"

Collecting like terms, you obtain that;

"2\\overrightarrow{AB} + 3\\overrightarrow{BC} + \\overrightarrow{CA} = 4\\overrightarrow{AL} \u2212 3\\overrightarrow{AL} \u2212 \\overrightarrow{AL} + 3\\overrightarrow{LC} \u2212 \\overrightarrow{LC} = \\overrightarrow{LC} \u2212\\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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