Question

In a triangle ABC, angle ABC = 45, prove that AC^2 = AB^2 + BC^2 - 4 ar (triangle ABC)

Answer 1

Please take a look at the attached figure.

Angle ABC measures 45° (given). Hence angle OAB will also be 45°
As a result x = h

Now the area of triangle ABC = 1/2(BC)(h) = 1/2(BC)(x) ; since h = x,
Or we can write BC = 2(Area of Triangle ABC)/x ----------------(i)

We can also write AC² = h² + y²
But y is nothing but (BC - x)

So we can write AC² = h² + (BC  - x)²

Simplifying this we get,
AC² = h² + BC² + x² - 2.x.BC
Now x² + h² = AB² (by Pythagoras theorem for triangle AOB)

Hence AC² = AB² + BC² -2(x)(BC)
Substituting the value of BC obtained from (i), we get
AC² = AB² + BC² - 2(x)(2Area of Triangle ABC)/x

Simplifying this we get:
AC² = AB² + BC² -4.Area of Triangle ABC

Hence proved.

Answer 2

It is the required answer. I hope it will help u