Question

In the given figure ABCD is a parallelogram E is mid-point of AB and CE bisects ZBCD. Prove that DA D AE AD 1) DE bisects ADC 18) DEC = 90 AL A B E​

Answer 1

Step-by-step explanation:

Given : || gm ABCD in which E is mid-point of AB and CE bisects ∠BCD.To Prove : (i) AE = AD(ii) DE bisects ∠ADC(iii) ∠DEC=90∘Const. Join DEProof : (i) AB || CD            (Given)and CE bisects it.∴         ∠1=∠3 (alternte ∠s)        ...(i)But     ∠1=∠2  (Given)      ...(ii)From (i) & (ii)              ∠2=∠3∴       BC=BE                    (sides opp. to equal angles)But        BC=AD (opp. sides of || gm)and        BE=AE (Given)              AD=AE∴           ∠4=∠5  (∠s opp. to equal sides)But         ∠5=∠6         (alternate ∠s)⇒           ∠4=∠6∴  DE bisec