Question
$\mathrm{ABCD}$ is a trapezium in which $\mathrm{AB}$ II $\mathrm{DC}$ and its diagonals intersect each other at the point $\mathrm{O} .$ Showthat $\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$
Step 1
We also need to draw a line EO parallel to AB and DC. Show more…
Show all steps
Your feedback will help us improve your experience
Raushan Kumar and 71 other Geometry educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
$$ \begin{array}{l} \text { The diagonals of a quadrilateral ABCD intersect each other at the point } \mathrm{O} \text { such that }\\ \frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}} \cdot \text { Show that } \mathrm{ABCD} \text { is a trapezium. } \end{array} $$
Triangles
$\quad$ Similar Figures
Diagonals AC and BD of a trapezium ABCD with $\mathrm{AB}$ II DC intersect each other at the point $\mathrm{O}$. Using a similarity criterion for two triangles, show that $\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$.
Similarity of Triangles
If $A B C D$ is a trapezium, $A C$ and $B D$ are the diagonals intersecting each other at point $O$. Then $\mathrm{AC}: \mathrm{BD}=$ (1) $\mathrm{AB}: \mathrm{CD}$ (2) $\mathrm{AB}+\mathrm{AD}: \mathrm{DC}+\mathrm{BC}$ (3) $\mathrm{AO}^{2}: \mathrm{OB}^{2}$ (4) $\mathrm{AO}-\mathrm{OC}: \mathrm{OB}-\mathrm{OD}$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD