Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangent QA and QB are equal.

Hint: The two tangents from a common point onto the same circle are equal.
We have a tangent which is passing through Q and P, which we are going to use with the help of the properties of the tangent. We will prove the required. We are going to prove the tangent QP is equal to QA and as QP is tangent to the second circle, it will be equal to QB. Which indirectly proves what is required.

Complete step by step answer:
We are given that Q is a common tangent through P, which means there are three tangents given which are QA, QB, QP.
We cannot directly prove $QA = QB$. First, we have to prove $QA = QP$ then later prove $QB = QP$
And then we will be able to prove $QA = QB$.
First, we prove $QA = QP$. So
From Q, QA and QP are two tangents to the circle with centre O.
Therefore, we get
$QA = QP - (1)$
Similarly, from Q, QB and QP are two tangents to the circle with centre O.
Therefore, we get
 $QB = QP - (2)$
From the condition (1) and condition (2) from the above, we get
$QA = QB$
Hence proved.

Note: We should always remember that the tangents can be equal only if they have a circle with a common centre and tangents with a common point and with a circle of different centres are never equal in length-wise by the properties of tangents.