Suppose given an isosceles triangle with a leg measuring 5 in. Two lines are drawn through some point on the base, each parallel to one of the legs. Find the perimeter of the constructed parallelogram.

Let ΔABC be isosceles triangle with legs AB=BC=5 in. Draw an arbitrary point D on the base AC and spend two lines DE and DF such that DE || BC and DF || AB.You get the parallelogram BFDE. 1. Consider ΔAED. This triangle is isosceles, because ∠ADE≅∠ACB as corresponding angles at parallel lines. The legs of this triangle are AE=ED.2. Consider ΔDFC. This triangle is isosceles, because ∠FDC≅∠BAC as corresponding angles at parallel lines. The legs of this triangle are DF=FC.3. Find the perimeter of parallelogram BFDE. [tex]P_{BFDE}=BF+FD+ED+BE.[/tex]Since AE=ED and DF=FC, you have that[tex]P_{BFDE}=BF+FC+AE+BE=AB+BC=5+5=10\ in.[/tex]Answer: 10 in.