In a triangle ABC, a line is drawn from C which bisects AB at point D. Find the ratio of area of the triangles DBC and ABC.

Given : CD bisects AB, => AD = DB = $$\frac{8}{2}=4$$ cm

To find : $$\frac{ar(\triangle DBC)}{ar(\triangle ABC)}=?$$

Solution : Clearly $$\triangle$$ ABC is a right angled triangle, $$\because (10)^2=(8)^2+(6)^2$$

Thus, AC is the hypotenuse and $$\triangle$$ ABC is right angled at B.

=> AB = 8 cm is the height of triangle

$$\therefore$$ $$\frac{ar(\triangle DBC)}{ar(\triangle ABC)}=\frac{\frac{1}{2}\times(DB)\times(BC)}{\frac{1}{2}\times(AB)\times(BC)}$$

= $$\frac{4\times6}{8\times6}=\frac{1}{2}$$

=> Ans - (C)