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If ABCD is a parallelogram, then prove that
ЁЭСОЁЭСЯ (ΔЁЭР┤ЁЭР╡ЁЭР╖) = ЁЭСОЁЭСЯ (ΔЁЭР╡ЁЭР╢ЁЭР╖) = ЁЭСОЁЭСЯ (ΔЁЭР┤ЁЭР╡ЁЭР╢) = ЁЭСОЁЭСЯ (ΔЁЭР┤ЁЭР╢ЁЭР╖) = `1/2` ЁЭСОЁЭСЯ (||ЁЭСФЁЭСЪ ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖) .
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Given: ABCDis a parallelogram
To prove : area (ΔЁЭР┤ЁЭР╡ЁЭР╖) = ЁЭСОЁЭСЯ (ΔAЁЭР╡ЁЭР╢) = are (Δ ACD)
= `1/2` ЁЭСОЁЭСЯ (||ЁЭСФЁЭСЪ ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖)
Proof: we know that diagonals of a parallelogram divides it into two equilaterals.
Since, AC is the diagonal.
Then, ЁЭСОЁЭСЯ (ΔЁЭР┤ЁЭР╡ЁЭР╢) = (Δ ACD) = `1/2` ЁЭСОЁЭСЯ (||ЁЭСФЁЭСЪ ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖)............ (1)
Since, BD is the diagonal
Then, ЁЭСОЁЭСЯ (ΔЁЭР┤ЁЭР╡ЁЭР╢) = ЁЭСОЁЭСЯ (ΔЁЭР╡ЁЭР╢ЁЭР╖) = `1/2` ЁЭСОЁЭСЯ (||ЁЭСФЁЭСЪ ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖)............ (2)
Compare equation (1) and (2)
∴ ЁЭСОЁЭСЯ (ΔЁЭР┤ЁЭР╡ЁЭР╢) = ЁЭСОЁЭСЯ (ΔЁЭР┤ЁЭР╢ЁЭР╖)
= ЁЭСОЁЭСЯ (ΔЁЭР┤ЁЭР╡ЁЭР╖) = ЁЭСОЁЭСЯ (ΔЁЭР╡ЁЭР╢ЁЭР╖) = `1/2` ЁЭСОЁЭСЯ (||ЁЭСФЁЭСЪ ЁЭР┤ЁЭР╡ЁЭР╢ЁЭР╖)
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