The vertices of a triangle ABC are A(1,2), B(4,6) C(6,14). AD is the bisector of the angle A meets BC at D. Find the coordinates of the point D.

As the angle bisector divides the sides in a proportional ratio.

So $\frac{AB}{AC}=\frac{BD}{CD}$
Hence AB = $\sqrt{(4-1{)}^2 + (6-2{)}^2} =5$
And AC = $\sqrt{(6-1{)}^2 +(14-2{)}^2}=13$
So AB/AC = 5/13
So point D(x,y) divides BC in the ratio of 5:13
Hence from internal division formula
x = $\frac{5\times 6 + 13\times 4}{5+13}, y =\frac{5\times 14 +13\times 6}{5+13}$
So x = 82/18 , y = 148/18
So coordinate of D is (82/18, 148/18)