Question

Given: In Circle O, BE≅BD
BE⊥OA
BD⊥OC
Prove: Arc AB≅ Arc BC

Answer 1

Given: In Circle O, BE≅BD

BE⊥OA

BD⊥OC

To Prove:

Arc AB≅ Arc BC

Solution:

To prove that two arcs are congruent, we need to prove that the angles subtended by the arcs on the center of the circle are equal.

For this, let us consider the triangles:

$\triangle BOE$ and $\triangle BOD$.

Given that: BE⊥OA  i.e. $\angle BEO =90^\circ$

BD⊥OC i.e. $\angle BDO =90^\circ$

1. $\angle BEO =\angle BDO =90^\circ$

2. Sides BE $\cong$ BD

3. Side BO is common to both the triangles.

So, $\triangle BOE$ $\cong$ $\triangle BOD$

$\therefore$ we can say that all the angles and sides of $\triangle BOE$ and $\triangle BOD$ must also be congruent.

$\therefore$ $\angle BOE\cong \angle BOD$ .

These are the angles subtended by the arcs AB and BC respectively on the center of the circle.

So, we can say that  Arc AB≅ Arc BC.

Hence proved.