math
posted by avadhesh on .
in a triangle ABC it is known that AB=AC. Suppose D is the mid point of AC and BD=BC=2. Then the area of the triangle ABC is

let AD = DC = x
then AB = 2x
in triangle ABC , angle B = angle C =Ø
in triangle BCD , angle C = angle BDC = Ø
2 angles of one are equal to 2 angles of the other, so they are similar
then:
2/x = 2x/2
2x^2 = 4
x^2=2
x=√2
So triangle ABC has equal sides of 2√2 and a base of 2
which is similar to
√2 : √2 : 1
Using the cosine law we can find angle A
1^2 = √2^2 + √2^2  2(√2)(√2)cosA
4cosA = 3
cosA = 3/4
then sinA = √7/4
area ABC = (1/2)(2e√2)(2√2)(√7/4)= √7
check my arithmetic, should have written it out first.