Two triangles ΔABC and ΔDBC are on the same base BC and on the same side of BC in which ∠A=∠D=90. If CA and BD meet each other at E, then show that AE×EC=BE×ED.
Given triangles ABC and DBC are n the same base BC.Consider Δ′s ABC and DBC
∠A=∠D=900 (given)
∠AEB=∠DEC (vertically opposite angles are equal )
Hence,
ΔABC∼ΔDBC (AA similarity theorem)
⇒AEDE=BECE∴AE×EC=BE×ED
Hence, proved.