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In the given figure, AB||CD, angleA=90^(...

In the given figure, `AB||CD, angleA=90^(@)` and `angleAEC=40^(@)`. Find `angleECD`.

Text Solution

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Through C draw `CF||AE`, cutting AB at G.
Now, `CF||AE` and EC is the transversal.
`:. angleECG=angleAEC=40^(@)` [alternate int. `angles`].
Now, `CF||AE` and AGB is a trasversal.
`:. angleFGB=angleEAG=90^(@)` [corresponding `angles`]
Again, `AB||CD` and GC is the transversal.
`:. angleGCD=angleFGB=90^(@)` [corresponding `angles`]
`:. angleECD=angleECG+angleGCD=(40^(@)+90^(@))=130^(@)`.
Hence, `angleECD=130^(@)`.
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