ABCD is a trapezium in which AB is parallel to DC and its diagonals intersect each other at point O .Show that AO/BO=CO/DO
google is your friend. Go to
http://www.authorstream.com/Presentation/ramarao.velury-1400926-7similartriangles-rev1/
and skip to page 12. The problem is explained.
Answer
To prove that AO/BO=CO/DO in the trapezium ABCD, we will use the property of similar triangles.
Let's consider triangles AOB and COD. By the property of similar triangles, we have:
AO/CO = AB/CD (Since AB || DC)
and
BO/DO = AB/CD (Since AB || DC)
Now, to prove that AO/BO=CO/DO, we can cross multiply the two equations:
(AO/CO) * (BO/DO) = (AB/CD) * (AB/CD)
Simplifying the equation further, we get:
(AO * BO) / (CO * DO) = (AB * AB) / (CD * CD)
Now, let's observe that triangles AOB and COD are similar, so the ratios of their corresponding sides are equal:
AB/CD = AO/CO = BO/DO
Hence, we can substitute the equal ratios in the equation:
(AO * BO) / (CO * DO) = (AO * AO) / (CO * CO)
Now, cross-multiplying the equation again, we get:
(AO * BO * CO * CO) = (AO * AO * CO * DO)
Cancelling the common terms on both sides, we get:
(AO * BO) = (AO * AO)
Dividing both sides by AO, we get:
AO/BO = AO/AO
Simplifying further, we get:
AO/BO = 1
Hence, we have proved that AO/BO=CO/DO in the trapezium ABCD.