Step-by-step explanation:
(i) Here in ∆ AED and ∆ CEB ,
- ang . DAE = ang. BCE
- AD = BC
- ang. AED = ang. BEC
Hence by AAS congruence congruence condition ,
- ∆AED [tex]\cong [/tex] ∆CEB
Also ,
- AE = CE . . . . . (i)
- BE = DE . . . . . (ii)
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(ii) In ∆ AEB and ∆CED ,
- AE = CE
- BE = DE
- ang. DEC = ang. AEB ( vertically opposite angles)
Hence by SAS condition ,
- ∆AEB [tex]\cong [/tex] ∆CED
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(iii) Here ,
Since these two angles are alternate interior Angles and equal . Hence the lines will be parallel . Also since ∆AEB is congruent to ∆CED , the corresponding sides will be equal . Hence AB = CD .
Hence Proved !
