Let $C$ be a circle with centre 'O' and $HK$ is the chord of contact of tangents drawn from a point $A$. $OA$ intersects the circle 'C' at $P$ and $Q$ and $B$ is the midpoint of $HK$, then:
AB is the harmonic mean of AP and AQ
OA is the arithmetic mean of AP and AQ
(AK)^2 = (OA)(AB)
AB is the geometric mean of AP and AQ
Step-by-Step Solution
Key Concept: The harmonic mean relationship emerges from equating two expressions derived from the angle bisector and power of a point.
Step 1: Relate $OA$ with $AP$ and $AQ$.
Let $R$ be the radius of the circle with center $O$. Therefore, $OP = OQ = R$.
Since $P$ and $Q$ are the points where the line segment $OA$ intersects the circle, and $A$ is an external point, $P$ lies between $A$ and $O$, while $O$ lies between $P$ and $Q$.
We can express $AP$ and $AQ$ in terms of $OA$ and $R$:
$$ AP = OA - OP = OA - R $$
$$ AQ = OA + OQ = OA + R $$
Adding these two equations, we get:
$$ AP + AQ = (OA - R) + (OA + R) = 2OA $$
Therefore, $OA$ is the arithmetic mean of $AP$ and $AQ$:
$$ OA = \frac{AP + AQ}{2} $$
This validates Option 2.
Step 2: Establish the relationship between $AK$, $OA$, and $AB$.
$K$ is a point of tangency, so the radius $OK$ is perpendicular to the tangent $AK$. Thus, $\triangle OKA$ is a right-angled triangle at $K$.
$HK$ is the chord of contact from point $A$. The line segment $OA$, which connects the external point $A$ to the center $O$, is perpendicular to the chord of contact $HK$. Since $B$ is the midpoint of $HK$, $B$ lies on $OA$ and $OA \perp HK$. Therefore, $\angle KBA = 90^\circ$.
Consider the right-angled triangles $\triangle ABK$ (right-angled at $B$) and $\triangle OKA$ (right-angled at $K$).
Both triangles share the common angle $\angle OAK$.
Thus, the triangles are similar: $\triangle ABK \sim \triangle OKA$.
From the property of similar triangles, the ratio of corresponding sides is equal:
$$ \frac{AB}{AK} = \frac{AK}{OA} = \frac{BK}{OK} $$
From the first equality, we derive the relation:
$$ (AK)^2 = (OA)(AB) $$
This validates Option 3.
Step 3: Relate $AK$, $AP$, and $AQ$ using the Power of a Point Theorem.
The Power of a Point Theorem states that for an external point $A$, the square of the length of the tangent segment from $A$ to the circle ($AK^2$) is equal to the product of the lengths of the segments of any secant line from $A$ through the circle ($AP \cdot AQ$).
Thus, we have:
$$ (AK)^2 = (AP)(AQ) $$
Step 4: Combine the results to find the relationship between $AB$, $AP$, and $AQ$.
From Step 2, we established $(AK)^2 = (OA)(AB)$.
From Step 3, we established $(AK)^2 = (AP)(AQ)$.
Equating these two expressions for $(AK)^2$:
$$ (OA)(AB) = (AP)(AQ) $$
Now, substitute the expression for $OA$ from Step 1 ($OA = \frac{AP + AQ}{2}$):
$$ \left(\frac{AP + AQ}{2}\right)(AB) = (AP)(AQ) $$
Solving for $AB$:
$$ AB = \frac{2(AP)(AQ)}{AP + AQ} $$
This formula shows that $AB$ is the harmonic mean of $AP$ and $AQ$. This validates Option 1.
Step 5: Conclude the correct options.
Based on the derivations in Step 1, Step 2, and Step 4:
Option 1: AB is the harmonic mean of AP and AQ (verified).
Option 2: OA is the arithmetic mean of AP and AQ (verified).
Option 3: $(AK)^2 = (OA)(AB)$ (verified).
Option 4: AB is the geometric mean of AP and AQ (not verified, as $AB$ is the harmonic mean).
The final answer is $\boxed{\text{1,2,3}}$.
Correct Answer: 1,2,3