Prove that the area of the triangle described on one side BC of a square ABCD as base is half the area of the similar triangle ACF produced on the diagonal AC as base.
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Solution
ABCD is a square with AC as its diagonal.
Then,
AC2=AB2+BC2=BC2+BC2=2BC2
⇒AC=√2BC
Also, it is given that ΔBCE ∼Δ ACF
Using the theorem, "The area of 2 similar triangles are in ratio of squares of the corresponding sides."
Area(ΔBCF)Area(ΔACF)=BC2AC2=BC2(√2BC)2=12
⇒ Area (ΔBCE)=12.
Area (ΔACF)
Hence, the area of ΔBCE is half of the area of ΔACF.