$△$ COB is an equilateral triangle and OD=OA.
Which of these statement is correct?

AB = DB

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OC = DO

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AB = DC

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DB = AB

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Solution

The correct option is C AB = DC
Consider $△$ COD and $△$ BOA
CO=OB (sides of an equilateral triangle)
$∠$ COD= $∠$ BOA (vertically opposite angles)
OD=OA (given )
Hence $△$ COD $≅$ $△$ BOA by SAS postulate.

So,AB=DC (corresponding parts of congruent triangles)

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Similar questions

(O A + O B + O C) > (A B + B C + C A)

(O A + O B + O C) > \frac{1}{2} (A B + B C + C A)

(O A + O B + O C) < \frac{1}{2} (A B + B C + C A)

A B

C D

^{'} O^{'}, O A = O B

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△ A O D ≅ △ B O C

If $O$ is a point within triangle $A B C$ , show that,

. O is any point in the interior of a triangle ABC. Prove that

1. AB + AC > OB + OC

2.AB + BC + CA > OA + OB + OC

3.OA + OB + OC > 1/2(AB + AC + BC)

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