If ABCD is a parallelogram, and X and Y be any two points on DC and AD respectively. Show that the triangle AXB and BYC are equal in area.

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$$\begin{gathered} \mathrm{Since} ∆\mathrm{AXB} \mathrm{and} \mathrm{parallelogram} \mathrm{ABCD} \mathrm{have} \mathrm{same} \mathrm{base} \mathrm{AB} \mathrm{and} \mathrm{between} \mathrm{same} \mathrm{parallels} \mathrm{AB} \mathrm{and} \mathrm{DC}, \mathrm{then} \\ \mathrm{area} \mathrm{of} ∆\mathrm{AXB} = \frac{1}{2}\left(\mathrm{area} \mathrm{of} {\parallel }^{\mathrm{gm} }\mathrm{ABCD}\right) ....\left(1\right) \\ \mathrm{Since} ∆\mathrm{BYC} \mathrm{and} \mathrm{parallelogram} \mathrm{ABCD} \mathrm{have} \mathrm{same} \mathrm{base} \mathrm{BCand} \mathrm{between} \mathrm{same} \mathrm{parallels} \mathrm{BC} \mathrm{and} \mathrm{AD}, \mathrm{then} \\ \mathrm{area} \mathrm{of} ∆\mathrm{BYC} = \frac{1}{2}\left(\mathrm{area} \mathrm{of} {\parallel }^{\mathrm{gm} }\mathrm{ABCD}\right) ....\left(2\right) \\ \mathrm{from} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ \mathrm{area} \mathrm{of} ∆\mathrm{AXB} = \mathrm{area} \mathrm{of} ∆\mathrm{BYC} \end{gathered}$$

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