A point O is taken inside a rhombus ABCD such that its distances from the vertices B and D are equal. Show that AOC is a straight line.

Rhombus ABCD with point O taken inside having equal distances from the vertices B and D is shown below:

In △ AOB and △ AOD,

⇒ AO = AO (Common side)

⇒ AB = AD (All sides of rhombus are equal)

⇒ OB = OD (O is equidistant from vertices B and D)

∴ ∆ AOB ≅ ∆ AOD (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠AOB = ∠AOD = x (let)

Since, BOD is a straight line.

∴ ∠AOB + ∠AOD = 180°

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180°}{2}$ = 90°

∴ ∠AOB = ∠AOD = 90°.

From figure,

∠BOC = ∠AOD = 90° (Vertically opposite angles are equal).

Since,

∠AOB + ∠BOC = 90° + 90° = 180°.

Thus, it can be said that points A, O and C lie on a straight line.

Hence, proved that AOC is a straight line.