Question

If in Fig 6.1, O is the point of intersection of two chords AB and CD such that OB = OD, then prove taht triangles OAC and ODB are similar (only use properties and theorems of grade 10 triangles)

Answer 1

Answer:

The proof is explained below.

Step-by-step explanation:

Given O is the point of intersection of two chords AB and CD such that OB = OD then we have to prove that triangles OAC and ODB are similar.

By chord intersection theorem which states that when two chords inside the circle intersect each other then the product of their segments are equal.

⇒ AO.OB=CO.OD

⇒ AO=CO    (∵OB=OD)

In triangle AOC,

∠ACO+∠AOC+∠CAO=180°

gives 2∠ACO+45°=180° ⇒ ∠ACO=67.5°         (∵AO=OC)

In triangle BOD,

∠BOD=∠COA=45°    ( vertically opposite angles)

∠BDO+∠BOD+∠DOB=180°

gives 2∠BDO+45°=180° ⇒ ∠BDO=67.5°

In triangle AOC and BOD  

∠ACO=∠BDO      (proved above)

∠BOD=∠COA      (vertically opposite)

Hence, by AA similarity, triangle AOC is similar to triangle BOD

Answer 2

Answer:

Correct option is

D

Isosceles and similar

Length of OB=OD and OC=OA

so it is isosceles

OA=OB=OC=OD=radius of circle

So it is similar.

Step-by-step explanation:

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