Question

$\triangle ACP \sim \triangle BCQ$
But how do you get $\triangle ACP \sim \triangle BQC$ ?

Answer 1

APxBQ = ACxAC or AP / AC = AC / BQ = BC / BQ as AC=BC ----------(1)

Interior angles A and B of triangle ABC are equal,
hence exterior angles CAP and CBQ are equal -----------------(2)

From (1) and (2),
(i) Two sides in triangles ACP and BQC are proportional 
(ii) Included angles of these proportional sides are equal

Hence the two triangles are similar.

Answer 2

ΔABC  is isosceles.
therefore, AB = AC.
AC²=BQ x AP
$\frac{AP}{AC}= \frac{AC}{BQ} \\ since, AC=AB \\ \frac{AP}{AC}= \frac{BC}{BQ}$
ang CAB = ang CBA (angle oppsite to equal sides)
⇒ang CAP = ang CBQ ---------------(1)
In ΔACP and Δ BQC,
$\frac{AP}{AC}=\frac{BC}{BQ}$
also, ang CAB = ang CBA (from (1))
hence,ΔCAP  similar to ΔCBQ (SAS)