Can someone help me with a geometry question...?

Question

A triangle ABC, Angle A=60. Prove that BC^2=AB^2 + AC^2 - (AB.AC) without using cosine rule.

Kindly help me.

Answer 1

In triangle ABC, draw a perpendicular BD from B to AC.
in triangle ABD, angle A = 60 degress(given)
angle BDA = BDC = 90 deg (by construction)
so Cos A = cos 60 = AD/BD
AD/AB = 1/2
AD = AB/2
in triangle BDC ,
BC^2 = BD^2 + DC^2(by pythogores theorem)
but BD^2 = AB^2 - AD^2 (from right angled triangle ADB)
and DC^2 =(AC -AD)^2 (since DC = AC - AD)
so BC^2 = AB^2 - AD^2 +(AC -AD)^2
= AB^2 - AD^2 + AC^2 +AD^2 - 2AC.AD
= AB^2 + AC^2 - 2AC. AD
substituting AD = AB/2
BC^2 = AB^2 + AC^2 - 2AC(AB/2)
BC^2 = AB^2 + AC^2 - AC(AB)

Answer 2

Okay, let's assume that AB > AC, so we extend AC and draw an equilateral triangle of sides AB, the new point being D. The altitude of this triangle is (1/2)√3 AB, so that the altitude of the triangle BCD is (1/2)√3 (AB - AC). Draw the right triangle BCE, where E lies on the line BD. The base of the right triangle, BE, is BD - (1/2)(AB - AC), but since BD = AB, BE = (1/2)(AB + AC). Using the Pythagorean, the square of the hypotenuse BC² = ((1/2)√3 (AB - AC))² + ((1/2)(AB + AC))² = AB² + AC² - (AB)(AC). QED.

Just pay close attention to the properties of the 30-60-90 right triangle to follow this proof.