Question

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

Answer 1

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,

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

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

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

Collecting like terms, you obtain that;

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