Question

In triangle abc angle acb=45° prove ac^2=ab^2+bc^2-4ar(∆abc)

Answer 1

Answer:

Step-by-step explanation:

Angle ABC measures 45° ....given

Therefore OAB will also 45°

As a result x = h

Now the area of triangle ABC = 1/2*(BC) (x).....

Therefore

BC = 2(AREA OF TRAIANGLE ABC) /x

Ac² = h² + y²

But y = (bc - y)

So we can write AC² = h² + (BC - x)²

Now, x² + h² = AB²

AC² = AB² + BC² - 2(x)(Area of triangle ABC)

BY SIMPLIFYING

AC² = AB² + BC² -4.Area of triangle ABC.......

HENCE PROVED

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