ABCD is a trapezium in which AB||CD and it's diagonals intersects each other at O. Show that AO/OB=CO/OD.guys plz helppppp....
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Given, AB is pallarel to CD. And its diagonal intersects at O.
To prove: AO/OB = CO/OD.
Construction : Draw a line EF parallel to AB & CD
⇒ Now, In triangle ADC,
EO is parallel to DC (bcoz EF is parallel to CD)
∴AE/DE = AO/CO.......→eq.1
Similarly, In triangle DBA
EO is parallel to AB (bcoz EF is parallel to AB)
∴ AE/DE = BO/DO......→eq.2
From eq. 1 and 2:-
AO/CO = BO/DO
∴ AO/BO = CO/DO
Hence , proved.
If my solution is correct...please give me brainliest!
To prove: AO/OB = CO/OD.
Construction : Draw a line EF parallel to AB & CD
⇒ Now, In triangle ADC,
EO is parallel to DC (bcoz EF is parallel to CD)
∴AE/DE = AO/CO.......→eq.1
Similarly, In triangle DBA
EO is parallel to AB (bcoz EF is parallel to AB)
∴ AE/DE = BO/DO......→eq.2
From eq. 1 and 2:-
AO/CO = BO/DO
∴ AO/BO = CO/DO
Hence , proved.
If my solution is correct...please give me brainliest!
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