Question

Line segment DB and line segment AE are medians. If BC=6y+10, AB=22+3y, CE=6x+12, ED=2x+60, then find the value of x and y, and the length of the segments.

Answer 1

`x=45.5,\ y=(280)/(3).`

Explanation:

Line segment DB and line segment AE are medians, then

• 
  `AD=DC=(1)/(2)AC;`
• 
  `BE=CE=(1)/(2)BC.`.

This gives you that
`6x+12=(6y+10)/(2).`

By the triangle midline theorem,

`DE=(1)/(2)BC,`

then

`2x+60=(22+3y)/(2).`

Solve the system of two equations:

`\left\{\begin{array}{l}6x+12=(6y+10)/(2)\\ \\2x+60=(22+3y)/(2)\end{array}\right.\Rightarrow \left\{\begin{array}{l}12x+24=6y+10\\ \\4x+120=22+3y\end{array}\right.\Rightarrow \left\{\begin{array}{l}6x-3y+7=0\\ \\4x-3y+98=0\end{array}\right.`

Subtract the second equation from the first one:

`6x-3y+7-(4x-3y+98)=0,\\ \\2x=98-7,\\ \\2x=91,\\ \\x=45.5.`

Then

`4\cdot 45.5-3y+98=0,\\ \\3y=280,\\ \\y=(280)/(3).`