Prove that 'the angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the center.' Apply this property to solve: If $PA$ and $PB$ are tangents from an external point $P$ to a circle with center $O$ such that $\angle APB = 70^\circ$, find $\angle OAB$.

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Proof of Theorem: Given: A circle with center $O$, an external point $P$, and two tangents $PA$ and $PB$ touching the circle at $A$ and $B$ respectively. To Prove: $\angle APB + \angle AOB = 180^\circ$ Proof: Consider the quadrilateral $PAOB$ formed by tangents $PA$, $PB$, and radii $OA$, $OB$. We know that the radius at the point of contact is perpendicular to the tangent. Therefore: $\angle OAP = 90^\circ$ $\angle OBP = 90^\circ$ The sum of all interior angles of a quadrilateral is $360^\circ$: $$\angle APB + \angle POB + \angle OBA 	ext{ (wait, quadrilateral angles:)} \angle APB + \angle AOB + \angle OAP + \angle OBP = 360^\circ$$ Substitute the known right angles: $$\angle APB + \angle AOB + 90^\circ + 90^\circ = 360^\circ$$ $$\angle APB + \angle AOB + 180^\circ = 360^\circ$$ $$\angle APB + \angle AOB = 360^\circ - 180^\circ = 180^\circ$$
Hence proved that $\angle APB$ and $\angle AOB$ are supplementary. Numerical Solution: Given: $\angle APB = 70^\circ$ $PA$ and $PB$ are tangents to the circle with center $O$. Step 1: Find $\angle AOB$
Using the proven property: $$\angle APB + \angle AOB = 180^\circ$$ $$70^\circ + \angle AOB = 180^\circ$$ $$\angle AOB = 180^\circ - 70^\circ = 110^\circ$$ Step 2: Consider $	riangle OAB$ In $	riangle OAB$, $OA = OB$ (Radii of the same circle). Therefore, $	riangle OAB$ is an isosceles triangle where $\angle OAB = \angle OBA$ (Angles opposite to equal sides are equal). Step 3: Calculate $\angle OAB$ The sum of angles in $	riangle OAB$ is $180^\circ$: $$\angle OAB + \angle OBA + \angle AOB = 180^\circ$$ Since $\angle OAB = \angle OBA$, we can write: $$\angle OAB + \angle OAB + 110^\circ = 180^\circ$$ $$2 \cdot \angle OAB + 110^\circ = 180^\circ$$ $$2 \cdot \angle OAB = 180^\circ - 110^\circ = 70^\circ$$ $$\angle OAB = \frac{70^\circ}{2} = 35^\circ$$ Answer: $$\angle OAB = 35^\circ$$