Question

ABCD is a trapezium, In which AB || DC and it's diagonals intersect each other at point O. Show that OA/OB = OC/OD​

Answer 1

Answer:

Step-by-step explanation:

Given parameters

ABCD is a trapezium where AB || DC and diagonals AC and BD intersect at O.

To prove

AOBO=CODO

Construction

Draw a line EF passing through O and also parallel to AB

Now, AB ll CD

By construction EF ll AB

∴ EF ll CD

Consider the ΔADC,

Where EO ll AB

According to basic proportionality theorem

AEED=AOOC ………………………………(1)

Now consider Δ ABD

where EO ll AB

According to basic proportionality theorem

AEED=BOOD ……………………………..(2)

From equation (1) and (2) we have

AOOC=BOOD

⇒ AOBO=OCOD

Hence the proof.

Answer 2

★ Gɪᴠᴇɴ :

• ABCD is a trapezium
• AB || DC
• Diagonals intersect each other at point O

☢ Tᴏ Fɪɴᴅ :

• Show that OA/OB = OC/OD

☯ Sᴏʟᴜᴛɪᴏɴ :

In ∆OAB, ∆OCD

↠ ∠AOB = ∠COD (∵ pair of vertically opp. angles)

↠ ∠OAB = ∠COD (∵ pair of alternate angles)

↠ ∠OBA = ∠ODC (∵ pair of alternate angles)

∴ ∆OAB ∼ ∆OCD (∵ AAA similarity)

❥ OA/OC = OB/OD ⇒ OA/OB = OC/OD