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Question

In the given figure, AB || CD, ∠BAE = 100° and ∠AEC = 30°. Find ∠DCE. Figure

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Solution

$\mathrm{Construction}:\phantom{\rule{0ex}{0ex}}D\mathrm{raw}GH\parallel FC\text{}\mathbf{\text{as shown in the book}}\mathrm{by}\mathrm{dotted}\mathrm{line}GH,\phantom{\rule{0ex}{0ex}}\mathrm{which}\mathrm{meets}AB\mathrm{at}A\mathrm{and}CD\mathrm{at}H.\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\angle GAE=\angle AEC=30°\left(\mathrm{alternate}\mathrm{angle}\text{s}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ $\mathrm{And}\angle GAB=\angle BAE-\angle GAE\phantom{\rule{0ex}{0ex}}=100°-30°\phantom{\rule{0ex}{0ex}}=70°\phantom{\rule{0ex}{0ex}}\mathrm{Now},AB\parallel CD\mathrm{and}GH \mathrm{is}\text{the}\mathrm{transversal}.\phantom{\rule{0ex}{0ex}}\angle DHA =\angle GAB=70°\left(\mathrm{corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Again}GH\parallel FC\text{and}DC\mathrm{is}\text{the}\mathrm{transversal}.\phantom{\rule{0ex}{0ex}}\therefore \angle DCE=\angle DHA =70°\left(\mathrm{corresponding}\mathrm{angles}\right)$

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