Question

In a AABC, LABC = 45°. Prove that (AC) = (AB)2 + (BC) - 4 ar (AABC).​

Answer 1

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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.

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Answer 2

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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.