AB||CD and EF||DA find AnglePDQ, AngleAED, AngleDEF
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AB//cd and de is a transversal
angle cdp=angleAED
AgleAED=34
angleAED+angleDEF+angle FEB=180[LINEAR PAIR]
34+DEF+78=180
112+DEF=180
DEF=68
DEF=PDQ=68
angle cdp=angleAED
AgleAED=34
angleAED+angleDEF+angle FEB=180[LINEAR PAIR]
34+DEF+78=180
112+DEF=180
DEF=68
DEF=PDQ=68
Arunshah320:
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