Prove that area of any triangle described on oneside of a square as base is one-half of the area of similar triangle described on the diagonal as base
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Square ABCDAC is diagonal of square ABCDTriangle ADM is constructed on base AD such that ADM is equilateralTriangle ACN is constructed on base AC such that ACN is equilateral
To prove - ar ADM = 1/2 ar ACN
Proof-
Since Triangles ADM and ACN are equilateral, they are similar ...........(1)
In triangle ADC
AD2 + DC2 = AC2 ....(Pythagoras Theoren)
=> 2AD2 = AC2 ...(Since, AD = DC as AD and DC are sides of square ABCD)
=> AD2 = AC2/2
=> AD = AC/√2 .................(2)
Since triangle ADM ~ triangle ACN (proved)
=> ar ADM/ ar ACN = AD2/AC2 = DM2 /CN2 = AM2/AN2
=>ar ADM/ ar ACN = AD2/AC 2 .............(3)
=> putting value of (2) in (3) we get
=> ar ADM/ ar ACN = (AC/√2) 2 /AC2
=> ar ADM/ ar ACN = AC 2 / 2 /AC2
=> ar ADM/ ar ACN = 1/2
=> 2ar ADM = ar ACN
=> ar ADM = 1/2 ar ACNHence, proved....