Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/331

Rh 18. The perimeter of an isosceles triangle is less than that of any other triangle of equal area standing on the same base.

Let ABC be an isosceles triangle; AQC any other triangle equal in area and standing on the same base AC.

Join BQ; then BQ is parallel to AC (I. 39).

And it will follow from 17 that the sum of AQ and QC is greater than the sum of AB and BC.

19. If a polygon he not equilateral a polygon may he found of the same number of sides, and equal in area, but having a less perimeter.

For, let CD, DE be two adjacent unequal sides of the polygon. Join CE. Through D draw a straight line parallel to CE. Bisect CE at L; from L draw a straight line at right angles to CE meeting the straight line drawn through D at K. Then by removing from the given polygon the triangle CDE and applying the triangle CKE we obtain a polygon having the same number of sides as the given polygon, and equal to it in area, but having a less perimeter (18).