Math, asked by beautygirl98765, 1 year ago

please solve que 10 please GUYS

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Answered by okaps
8
Given: Quadrilateral ABCD in which diagonals AC and BD intersects each other at O such that AO/BO = CO/DO.


To Prove: ABCD is a trapezium


Construction: Through O, draw line EO, where EO || AB, which meets AD at E.


Proof: In ΔDAB, we have

EO || AB

∴ DE/EA = DO/OB ...(i) [By using Basic Proportionality Theorem]

Also,  AO/BO = CO/DO (Given)

⇒ AO/CO = BO/DO

⇒ CO/AO = BO/DO

⇒ DO/OB = CO/AO ...(ii) 


From equation (i) and (ii), we get

DE/EA = CO/AO

Therefore, By using converse of Basic Proportionality Theorem, EO || DC also EO || AB

⇒ AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.

Hope it helped !

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Answered by creamiepie
5
follow the attachment above
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MAYAKASHYAP5101: A1 aNsWeR
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