How can AO/OC=BO/OD become AO/BO=CO/DO in grade 10 ncert book sum's 9th question's solution?
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Step-by-step explanation:
Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.
We have to prove, AO/BO = CO/DO
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In ΔADC, we have OE || DC
Therefore, By using Basic Proportionality Theorem
AE/ED = AO/CO ……………..(i)
Now, In ΔABD, OE || AB
Therefore, By using Basic Proportionality Theorem
DE/EA = DO/BO…………….(ii)
From equation (i) and (ii), we get,
AO/CO = BO/DO
⇒AO/BO = CO/DO
Hence, proved..
hope is helpful to u
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