In the figure, BA`|""|` ED and BC`|""|` EF. Show that `/_`ABC = `/_`DEF.

In the figure, BA || ED and BC || EF. Show /_ABC + /_DEF = 180^(@) .

In the figure, BA || ED and BC || EF. Show /_ABC + /_DEF = 180^(@) .

In the given figure, BA||ED and BC||EF . Show that angleABC+angleDEF=180^(@) .

In fig. BA||ED and BC||EF. Show that angleABC=angleDEF

In Fig. 6.12, BA || ED and BC || EF. Show that angle ABC = angle DEF (Hint: Produce DE to intersect BC at P (say)]

In Figure,|CF and BC|ED. Prove that /_ABC=/_FDE

In fig. BA||ED and BC||EF> show that angleABC+angleDEF=180^@

In the given figure , AB II ED and BC II EF , then the value of angleABC +angle DEF is :

If Figure,arms BA and BC OF /_ABC are respectively parallel to arms ED and EF or /_DEF. Prove that /_ABC=/_DEF