B is the midpoint of a AC and D is the midpoint of CE solve for BD given that BD = 2Y + 8 and AE = 6Y +12

Answer 1

Since B is the midpoint of AC, we can write the equation:

AC = 2AB

Since D is the midpoint of CE, we can write the equation:

CE = 2CD

We also know that AC = AE + EC.

Substituting the given values, we get:

2AB = 6Y + 12 + 2CD

Rearranging the equation, we get:

2CD = 2AB - 6Y - 12

Since B is the midpoint of AC, AB = BC.

Plugging this into the equation, we get:

2CD = 2BC - 6Y - 12

Since D is the midpoint of CE, CD = DE.

Plugging this into the equation, we get:

2DE = 2BC - 6Y - 12

Since AB + BC = AC, we can substitute AB = AC/2 and BC = AC/2 into the equation:

2DE = AC - 6Y - 12

Since AC = 2AB, we can substitute AC = 2(BC) into the equation:

2DE = 2(BC) - 6Y - 12

2DE = 2BC - 6Y - 12

But we already know that 2CD = 2BC - 6Y - 12.

Therefore, 2DE = 2CD.

This means that DE = CD.

Since D is the midpoint of CE, we know that EC = 2CD.

We can substitute EC = 2CD into the equation:

2DE = EC

DE = EC/2

Finally, since D is the midpoint of CE, we can write the equation:

DE = (CE + EC)/2

Substituting the given values, we get:

DE = (6Y + 12 + 2CD)/2

Substituting CD = DE, we get:

DE = (6Y + 12 + 2DE)/2

Multiplying both sides of the equation by 2, we get:

2DE = 6Y + 12 + 2DE

Simplifying the equation, we get:

DE = 3Y + 6

So, BD = 2(3Y + 6) + 8

BD = 6Y + 12 + 8

BD = 6Y + 20