(a) A man is walking from a town A to another town B at a speed of $$4\text{km/hr}$$. He started one hour before a bus starts. The bus travelling at $$12\text{km/hr}$$. The man on the way got into the bus and travels 2 hours and reached town B. What is the distance between town A and town B? (b) A point P is taken within a rhombus ABCD such that $$PA=PC$$. Show that B, P, D are collinear.

For part (a), the man walks $$4\times 1=4\text{km}$$ before the bus starts, and since the bus closes this gap at a relative speed of $$12-4=8\text{km/hr}$$, it takes $$\frac{4}{8}=0.5\text{hr}$$ to catch up, covering an additional $$4\times 0.5=2\text{km}$$ on foot; after boarding, the bus travels 2 more hours at 12km/hr, covering $$24\text{km}$$. So the total distance is $$4+2+24=30\text{km}$$. For part (b), using SSS congruence, $$\triangle APD \cong \triangle CPD$$ gives equal angles at P on one side, and $$\triangle PAB \cong \triangle PCB$$ gives equal angles at P on the other side; since all four angles around P sum to $$360^\circ$$, this forces the two angles at P along line BD to sum to $$180^\circ$$, meaning B, P, D lie on a straight line.