**Note - the problem I'm struggling with is how to calculate the area of APBQ (the last question)
Figure 1 on the right shows a right-angled triangle ABC where AB ＝ 1 cm, AC ＝ 2 cm, and angle BAC ＝ 90°. Triangle PAB is an isosceles triangle where AP ＝ AB and sides PA and BC are parallel. Assume point P is located opposite to point C with respect to line AB.
Answer the following questions.
〔Question 1 〕 Consider the case in Figure 1 where the magnitude of angle APB is a°. Find the magnitude of angle ACB in terms of a.
〔Question 2 〕 Figure 2 on the right shows the case in Figure 1 where a perpendicular line to side BC is drawn from vertex A. Let Q be the intersection of side BC and the perpendicular line.
Answer （１）and （２）.
（１） Prove triangle ABQ is similar to triangle CAQ.
（２） Calculate the area of quadrilateral APBQ.