Posted by **avadhesh** on Monday, November 5, 2012 at 5:22am.

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

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**Reiny**, Monday, November 5, 2012 at 7:54am
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.

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