If ABCD is a parallelogram, then prove that
ππ (Δπ΄π΅π·) = ππ (Δπ΅πΆπ·) = ππ (Δπ΄π΅πΆ) = ππ (Δπ΄πΆπ·) = `1/2` ππ (||ππ π΄π΅πΆπ·) .
Solution
Given: ABCDis a parallelogram
To prove : area (Δπ΄π΅π·) = ππ (ΔAπ΅πΆ) = are (Δ ACD)
= `1/2` ππ (||ππ π΄π΅πΆπ·)
Proof: we know that diagonals of a parallelogram divides it into two equilaterals.
Since, AC is the diagonal.
Then, ππ (Δπ΄π΅πΆ) = (Δ ACD) = `1/2` ππ (||ππ π΄π΅πΆπ·)............ (1)
Since, BD is the diagonal
Then, ππ (Δπ΄π΅πΆ) = ππ (Δπ΅πΆπ·) = `1/2` ππ (||ππ π΄π΅πΆπ·)............ (2)
Compare equation (1) and (2)
∴ ππ (Δπ΄π΅πΆ) = ππ (Δπ΄πΆπ·)
= ππ (Δπ΄π΅π·) = ππ (Δπ΅πΆπ·) = `1/2` ππ (||ππ π΄π΅πΆπ·)