In Fig. 15.89, ABCD is a ||ᵍᵐ. O is any point on AC. PQ||AB and LM||AD. Prove that:
ar(||ᵍᵐ DLOP) = ar(||ᵍᵐ BMOQ)
Answers
Given : ABCD is a ||ᵍᵐ. O is any point on AC. PQ||AB and LM||AD.
To Prove : ar(||ᵍᵐ DLOP) = ar(||ᵍᵐ BMOQ)
Proof :
Since, a diagonal of a parallelogram divides it into two triangles of equal area :
∴ ar (∆ADC) = ar (∆ABC)
ar (∆APO) + ar (|| gm DLOP) + ar (∆OLC)
ar (∆AOM) + ar (|| gm DLOP) + ar (∆OQC) …………….(1)
Since, AO and CO are diagonals of || gms AMOP and OQCL.
∴ ar (∆APO) = ar (∆AMO) ……………..(2)
ar(∆OLC) = ar(∆OQC) ………………..(3)
On Subtracting eq (2) from (3), we get
ar (|| gm DLOP) = ar (|| gm BMOQ).
Hence ar(||ᵍᵐ DLOP) = ar(||ᵍᵐ BMOQ) .
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Answer:
Step-by-step explanation:
ar (∆ADC) = ar (∆ABC)
ar (∆APO) + ar (|| gm DLOP) + ar (∆OLC)
ar (∆AOM) + ar (|| gm DLOP) + ar (∆OQC) …………….(1)
Since, AO and CO are diagonals of || gms AMOP and OQCL.
∴ ar (∆APO) = ar (∆AMO) ……………..(2)
ar(∆OLC) = ar(∆OQC) ………………..(3)
On Subtracting eq (2) from (3), we get
ar (|| gm DLOP) = ar (|| gm BMOQ)