Triangle ABC is isosceles with AC -BC and BD is the median to side AC. If DB is 2 cm greater than DC and the perimeter of triangle DBC is 70 cm find BC

Set up equations, and sketch a diagram to help you.

Let's call BC = y and CD = x.

We know that BD is 2 cm less than CD, so, BD = (x - 2) cm

Perimeter = BC + CD + BD = y + x + (x - 2) = 70

Simplify.

Then, if you know the cosine rule, you can use this:

Angle ABC = Angle DBC.

because BD cuts AC into two equal sides. Let this angle be theta.

x^2 = (2x)^2 + (x+2)^2 - 2(x)(x+2)cos\theta

x^2 = (y)^2 + (x+2)^2 - 2(y)(x+2)cos\theta

Rearrange both for cos theta:

x^2 = (2x)^2 + (x+2)^2 - 2(x)(x+2)cos\theta

cos \theta = \frac{x^2 - (2x)^2 - (x+2)^2}{-2(x)(x+2)}

cos \theta = \frac{x^2 - (2x)^2 - (y)^2}{-2(x)(y)}

Equate both;

\frac{x^2 - (2x)^2 - (x+2)^2}{-2(x)(x+2)} = \frac{x^2 - (2x)^2 - (y)^2}{-2(x)(y)}

now, you have got 2 equations with 2 unknown variables. Find y, which is the length of BC.

I hope it helped! :)