Question

the line segment joining the midpoint M andN of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC.prove that AD =BC

Answer 1

Construction:− Join CM and DM.In ∆CMN and ∆DMNMN=MN (Common)∠CNM=∠DNM=90° (MN Is perpendicular to DC)CN=DN (Since N is the mid point of DC)By SAS congruency∆CMN≅∆DMN Therefore,CM=DM (CPCT)∠CMN=∠DMN (CPCT)∠AMN=∠BMN=90 (Since MN is perpendicular to AB)So,∠AMN−∠CMN=∠BMN−∠DMN (Since ∠CMN=∠DMN )∠AMD=∠BMCIn ∆AMD and ∆BMCDM=CM (Proved above)∠AMD=∠BMC (Proved above)AM=BM (M is the mid point of AB)By SAS congruency∆AMD≅∆BMCTherefore,AD=BC (CPCT)Hence Proved