ANSWER PLEASE BERY TUFF
Answers
Explanation:
1) area of triangle AOB=area of triangle COP.
In Δ AOC and Δ COP
∠ AOC = ∠ COP (common angle)
OA = OP ( ∵ BP ║ AC and O is intersection point of these lines )
OB = OC ( ∵ BP ║ AC and O is intersection point of these lines )
∴ ar (Δ AOC) = ar (Δ COP) (using SAS theorem)
Hence it is proved that, area of triangle AOB=area of triangle COP.
2) area of quadrilateral ABCD = to area of triangle APD
Given:
ABCD is a quadrilateral
BP ║ AC
Proof:
In Δ ABC and Δ PAC
CP = CP (common base)
AC = AC (common side)
∴ ar (Δ ABC) = ar (Δ PAC) (using SS criteria)
adding ar (Δ ADC) on both sides, we get,
ar (Δ ADC) + ar (Δ ABC) = ar (Δ ADC) + ar (Δ PAC)
as ADC + ABC = ABCD
and ADC + PAC = APD
we have,
∴ ar (ABCD) = ar (Δ APD)
Hence it is proved that, area of quadrilateral ABCD = to area of triangle APD
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