The tangents drawn from an external point to a circle subtend equal angles with the chord joining the two points of contact.

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Solution

REF. Image.
Let PA & PB are tangents
$O P ⊥ P A$
$[∵$ Tangent at any point of circle is $⊥ e_{q}$
to radius through point of contact]
$∴ ∠ O A P = 90^{\circ}$
Since PB is tangent $\Rightarrow O B ⊥ P B \Rightarrow ∠ O B P = 90^{\circ}$
The Quadrilateral OAPB
$∠ O B P + ∠ A O B + ∠ O A P + ∠ A P B = 360^{\circ}$
$90^{\circ} + ∠ A O B + 90^{\circ} + ∠ A B P = 360^{\circ}$
$∠ A P B + ∠ A O B = 180^{\circ}$
$∴ P r o v e d$

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The tangents drawn from an external point to a circle subtend equal angles with the chord joining the two points of contact.

REF. Image.Let PA & PB are tangents OP⊥PA[∵ Tangent at any point of circle is ⊥eqto radius through point of contact]∴∠OAP=90∘Since PB is tangent ⇒OB⊥PB⇒∠OBP=90∘The Quadrilateral OAPB∠OBP+∠AOB+∠OAP+∠APB=360∘90∘+∠AOB+90∘+∠ABP=360∘∠APB+∠AOB=180∘∴Proved