Question

ad = db and af = fc. df = 14 and de = 10.
bc = ?

Explanation:

Step1: Identify Midline Property

Since \(AD = DB\) and \(AF = FC\), \(D\) and \(F\) are midpoints of \(AB\) and \(AC\) respectively. By the Midline Theorem (or Midsegment Theorem) in triangles, the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. So, \(DF\) is the midline of \(\triangle ABC\), meaning \(DF=\frac{1}{2}BC\).

Step2: Solve for \(BC\)

We know \(DF = 14\). From the midline relationship \(DF=\frac{1}{2}BC\), we can solve for \(BC\) by multiplying both sides by 2: \(BC = 2\times DF\). Substituting \(DF = 14\), we get \(BC=2\times14 = 28\).

Answer:

28