Question

MN is a diameter of a circle with centre ' O ' . If BD = CD , prove that ∠OAD = ∠OCD

Answer 1

Answer:

$\sf \: Proved \: \angle \: OAD \: = \angle \: OCD$

Step-by-step explanation:

Given:

Mn is diameter of circle having centre O

and BD = OD,

To prove that:

$\angle \: OAD \: = \angle \: OCD$

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,

$\angle \: OBD \: = \angle \: OCD$

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,

$\angle \: OBD \: = \angle \: OAB \\ \angle \: OAB \: = \angle \: OAD (same \: angles) \\ \angle \: OBD \: = \angle \: OAD$

Consider above prove as statement B

From Statement A & B we can say that

$\angle \: OAD \: = \angle \: OCD$

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