given : AD is ⊥to AB and DC, BC is ⊥ to AB and DC, angle EDC ≅ angle ECD, AD ≅ BC. prove: triangle ADE ≅ triangle BCE

Answer:Hence proved triangle ADE ≅ triangle BCE by Side Angle Side congruent property.Step-by-step explanation:Given:AD ⊥ AB [tex]\&[/tex] CDBC ⊥ AB [tex]\&[/tex] CDAD = BC∴ ∠ A = ∠ B = ∠ C = ∠ D =90°∠ EDC = ∠ ECDSolution∠ C = ∠ BCE + ∠ ECD⇒ equation 1∠ D = ∠ ADE + ∠ EDC⇒ equation 2∠ C = ∠ D (given)Substituting equation 1 and 2 in above equation we get∠ BCE + ∠ ECD = ∠ ADE + ∠ EDCBut ∠ EDC = ∠ ECD (given)∴ ∠ ADE = ∠ BCE ED = EC (∵ base angles are same triangle is isosceles triangle)Now, In Δ ADE and Δ BCEAD =BC ∠ ADE = ∠ BDEED = EC∴ By Side Angle Side congruent propertyΔ ADE ≅ Δ BCE