# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/४१४

108${\displaystyle {\tfrac {1}{2}}}$.[1] The two diagonals (of the oblong of reference constructed with the aid of the given bījas) become the two (equal) sides of the (required) isosceles triangle. The base (of the oblong of reference), multiplied by two, becomes the base (of the required triangle). The perpendicular-side (of the oblong of reference) is the perpendicular (of the required triangle from the apex to the base thereof). The area (of the required triangle) is the area (of the oblong of reference).
108${\displaystyle {\tfrac {1}{2}}}$.^  The rationale of the rule may be made out thus:- Let ABCD be an oblong and let AD be produced to E so that AD=DE. Join EC. It will be seen that ACE is an isosceles triangle whose equal sides are equal to the diagonals of the oblong and whose area is equal to that of the oblong.