Question

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point ‘O’. Show that \(\frac{AO}{BO}=\frac{CO}{DO}\).

Answer 1

Given: In trapezium □ ABCD, AB || CD. 

Diagonals AC, BD intersect at O. 

R.T.P: \(\frac{AO}{BO}=\frac{CO}{DO}\)

Construction: 

Draw a line EF passing through the point ‘O’ and parallel to CD and AB. 

Proof: In △ACD, EO || CD 

∴ \(\frac{AO}{CO}=\frac{AE}{DE}\) …….. (1) 

[∵ line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem]

 In △ABD, EO || AB 

Hence, \(\frac{DE}{AE}=\frac{DO}{BO}\)

[∵ Basic proportionality theorem]

\(\frac{BO}{DO}=\frac{AE}{ED}\) …….. (2) [∵ Invertendo] 

From (1) and (2)

\(\frac{AO}{CO}=\frac{BO}{DO}\)

⇒ \(\frac{AO}{BO}=\frac{CO}{DO}\) [∵ Alternendo]