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The medians of Delta ABC intersect at po...

The medians of `Delta ABC` intersect at point G. Prove that:
area of `Delta AGB = (1)/(3) xx " area of " Delta ABC`

Text Solution

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Let D is the mid-point of BC.
In `Delta ABC`
D is the mid-point of BC
`:.` area of `Delta ABD = " area of " Delta ADC`
(`.:` Median divides the triangle into two equal area)..(1)
In `Delta GBC`
D is the mid-point of BC
`:.` area of `DeltaGBD = " area of " Delta GCD`
(`.:` median divides the triangle into two equal area)..(2)
Subtract (2) from (1), we get
area of `Delta ABD - " area of " DeltaGBD = " area of "DeltaACD - " area of "Delta GCD`
`rArr` area of `DeltaABG = " area of " Delta ACG`....(3)
Similarly, we can prove that
area of `DeltaACG = " area of " DeltaBCG`
(`:.` median divides the triangle into two equal area)...(4)
From (3) and (4), we get
area of `DeltaABG = " area of "DeltaACG = " area of " DeltaBCG`
`:.` area of `Delta ABG = (1)/(3) xx " area of " Delta ABC`
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