Question

related to Circle lesson

Answer 1

$\Huge{\boxed{\red{\mathbb{ANSWER}}}}$

$\bold{\underline{Given\:that,}}$

∠ABP = 60°

$\bold{\underline{To\:show,}}$

△PAB ≅ △AOB

In△PAB and △AOB,

AB = BA (common)

∠QBA = ∠BAP (alternate interior angles)

∠QAB = ∠ABP (alternate interior angles)

So, PAB ≅ △AOB.