Answer:
[tex] \sf \: Proved \: \angle \: OAD \: = \angle \: OCD[/tex]
Step-by-step explanation:
Given:
Mn is diameter of circle having centre O
and BD = OD,
To prove that:
[tex]\angle \: OAD \: = \angle \: OCD[/tex]
Solution:
Join the points O and B and draw OB,
On joining the line,
in ∆OCD and ∆OBD,
OC =OB → (Radius of same circle)
BD =CD → (from given)
OD =OD → (Common side in both the triangles)
Hence ∆OCD and ∆OBD are congruent from SSS property.
so we can say that,
[tex]\angle \: OBD \: = \angle \: OCD[/tex]
Consider above prove as statement A
Corresponding angles of congruent traingle.
in ∆ OAB,
OA = OB (radius of same circle)
hence ∆OAB is an isosceles traingle.
We know that opposite angle of isosceles traingle are always equal. hence,
[tex]\angle \: OBD \: = \angle \: OAB \\ \angle \: OAB \: = \angle \: OAD (same \: angles) \\ \angle \: OBD \: = \angle \: OAD[/tex]
Consider above prove as statement B
From Statement A & B we can say that
[tex]\angle \: OAD \: = \angle \: OCD[/tex]
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