Answer 1

Since x and y are the midpoint AC and AB respectively

∴ XY ll BC

Clearly, triangles BYC and BXC are on the same base BC and between the same parallels
XY and BC

  ∴ area (ΔBYC)  = area (BXC)

⇒  area  (ΔBYC) = ar (ΔBOC) = ar (ΔBXC) - ar (BOC)

⇒  ar (ΔBOY) = ar (ΔCOX)

⇒ ar ( BOY) + ar (XOY) = ar (ΔCOX) + ar (ΔXOY)

⇒ ar  (ΔBXY = ar (ΔCXY)

We observe that the quadrilateral XYAP and XYAQ are on the same base XY and between
the same parallel XY and PQ.

  ∴ area (quad XYAP ) ar (quad XYPA)        ....(2)

Adding (1) and (2), we get

ar (ΔBXY) + ar (quad  XYAP) = ar (CXY) + ar (quad XYQA)

⇒ ar (ΔABP) = ar (ΔACQ)