Notifications
Clear all
Triangles
1
Posts
2
Users
0
Likes
443
Views
0
03/06/2021 12:10 pm
Topic starter
The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.
1 Answer
0
03/06/2021 12:11 pm
Given, Quadrilateral ABCD where AC and BD intersects each other at O such that,
AO/BO = CO/DO.
Prove that, ABCD is a trapezium
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In ΔDAB, EO || AB
Therefore, By using Basic Proportionality Theorem
DE/EA = DO/OB ……………………(i)
Also, given,
AO/BO = CO/DO
⇒ AO/CO = BO/DO
⇒ CO/AO = DO/BO
⇒ 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.
This post was modified 4 years ago by Raavi Tiwari