In ∆ABC, angle abc = 45°.prove that AC² = AB² + BC² - 4ar(∆abc

1.	In ∆ABC, angle abc = 45°.prove that AC² = AB² + BC² - 4ar(∆abc).
Answer» Angle ABC measures 45° (given). Hence angle OAB will also be 45°As a result x = hNow the area of triangle ABC = 1/2(BC)(h) = 1/2(BC)(x) ; since h = x,\xa0Or we can write BC = 2(Area of Triangle ABC)/x ----------------(i)We can also write AC² = h² + y²But y is nothing but (BC - x)So we can write AC² = h² + (BC - x)²Simplifying this we get,AC² = h² + BC² + x² - 2.x.BCNow 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 getAC² = AB² + BC² - 2(x)(2Area of Triangle ABC)/xSimplifying this we get:AC² = AB² + BC² -4.Area of Triangle ABC

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