Two circles touch each other externally at point P. Seg AB is a common tangent touching the circles at A and B. The common tangent at P meets AB at Q. Prove that (i) Q is the midpoint of AB (ii) \(\angle APB = 90^\circ\).

Answer: QA = QP = QB, so Q is the midpoint of AB and \(\angle APB = 90^\circ\) (proved below)

[Figure: two circles touching at P; direct common tangent touches them at A and B; tangent at P meets AB at Q; seg PA and seg PB drawn.]

(i) QA and QP are tangent segments from Q to the first circle \(\therefore QA = QP\). QB and QP are tangent segments from Q to the second circle \(\therefore QB = QP\). Hence QA = QB and Q is the midpoint of AB.

(ii) In \(\Delta QAP\), QA = QP \(\therefore \angle QAP = \angle QPA = x\) (say). In \(\Delta QBP\), QB = QP \(\therefore \angle QBP = \angle QPB = y\) (say).

\(\angle APB = \angle QPA + \angle QPB = x + y\).

In \(\Delta APB\): \(\angle PAB + \angle PBA + \angle APB = 180^\circ\) \(\therefore x + y + (x + y) = 180^\circ\) \(\therefore x + y = 90^\circ\).

\(\therefore \angle APB = 90^\circ\). Hence proved.