Give the proof of herons formula
Answers
Geometrical Proof of Heron’s Formula
(From Heath’s History of Greek Mathematics, Volume2)
Area of a triangle = sqrt[s(s-a)(s-b)(s-c)], where s = (a+b+c)/2
The triangle is ABC. Draw the inscribed circle, touching the sides at D, E and F, and having its center at O.
Since OD = OE = OF, area ABC = area AOB + area BOC + area COA,
2.area ABC = p.OD, where p = a+b+c. (a = (length of) BC, etc.)
Extend CB to H, so that BH = AF. Then CH = p/2 = s. (since BD = BF, etc.)
Thus (area ABC)2 = CH2.OD2.
Draw OL at right angles to OC cutting BC in K, and BL at right angles to BC meeting OL in L. Join CL.
Then, since each of the angles COL, CBL is right, COBL is a quadrilateral in a circle.
Therefore, angle COB + angle CLB = 180 degrees.
But angle COB + angle AOF = 180 degrees, because AO bisects angle FOE, etc., so
Angle AOF = angle CLB
Therefore, the right-angled triangles AOF, CLB are similar, and
BC:BL = AF:FO = BH:OD
CB:BH = BL:OD = BK:KD
And from CB/BH = BK/KD, adding one to each side, CH/HB = BD/DK, or
CH:HB = BD:DK
It follows that
CH2:CH.HB = BD.DC:CD.DK = BD.DC:OD2 (since angle COK = 90 degrees)
Therefore (area ABC)2 = CH2.OD2 = CH.HB.BD.DC = s(s-a)(s-b)(s-c).Step-by-step explanation:
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