ABCD is a trapezium in which AB||DC and its diagonals intersect each other at point o. Show that: AO/BO= CO/DO
Draw the diagram of ur own-
Answers
Given: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.
To Prove: ar(Δ AOD) = ar(Δ BOC).
Proof: Δ ABD and Δ ABC are on the same base AB and between the same parallels AB and DC.
∴ ar(Δ ABD) = ar(Δ ABC) two triangles on the same base and between the same parallels are equal in area.
Subtract ar(Δ AOB) both sides,
⇒ ar(Δ ABD) - ar(Δ AOB) = ar(Δ ABC) - ar(Δ AOB)
⇒ ar(Δ AOD) = ar(Δ BOC).
Hope it will helps you...
Answer:
Given parameters
ABCD is a trapezium where AB || DC and diagonals AC and BD intersect at O.
To prove
AO/BO=CO/DO
Construction
Draw a line EF passing through O and also parallel to AB
Now, AB ll CD
By construction EF ll AB
∴ EF ll CD
Consider the ΔADC,
Where EO ll AB
According to basic proportionality theorem
AE/ED=AO/OC ………………………………(1)
Now consider Δ ABD
where EO ll AB
According to basic proportionality theorem
AE/ED=BO/OD ……………………………..(2)
From equation (1) and (2) we have
AO/OC=BO/OD
⇒ AO/BO=OC/OD
Hence the proof.
Step-by-step explanation:
hope it helps you
