Let ΔABC is inscribed in a circle of radius r. A perpendicular E is drawn of AC from the vertex B of triangle which meets AC at D and circle at E.
Let base of triangle AC=2x
and height of triangle BD=y
Then, area of triangle A=12×2x×y=xy ....(i)
∵AC and BE are chords of circle.
So, AD×DC=BD×DE
∵ radius of circle =r
DE=BE−DB=2r−y]
⇒x×x=y×(2r−y)
⇒x2=2ry−y2
⇒x=√2ry−y2 ....(ii)
From equation (i) and (ii),
A=√2ry−y2y
⇒A2=(2ry−y2)y2=2ry3−y4
⇒A2=2ry3−y4
Diff. w.r.t. y,
2AdAdy=6ry2−4y3 ....(iii)
For maximum value of A,dAdy=0
∴dAdy=0
⇒6ry2−4y3=0 [∵A≠0]
⇒2hy2(3r−2y)=0
3r=2y [∵y≠0]
⇒y=32r
Diff. (iii) again w.r.t. y.
2dAdy.dAdy+2A.d2Ady2=12ry−12y2
⇒2(dAdy)2+2Ad2Ady2=12ry−12y2 ...(iv)
Putting in equation (iv) dAdy=0 and y=32r
2A.d2Ady2=12r×32r−12×94r2
=18r2−27r2=−−r2<0
∴d2Ady2<0 (∵2A>0)
So, area of triangle will be maximum
When y=32r then x=√2r×32r−94r2
⇒x=√3r2−94r2
=√12r2−9r24=√3r24
x=√3r2
In right angle ΔABC,
tanA=BDAD=yx=32r√3r2=√3
⇒tanA=tan60∘
⇒A=60∘
Similarly tanC=BDAD=yx=32r√3r2=√3
⇒tanC=tan60∘
⇒C=60∘
∴∠ABC=60∘
Hence, ΔABC is an equilateral triangle.
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