In parallelogram ABCD, AB = 14 cm and AD = 5 cm. The bisector of A meets side DC at E. What is EC?

Question 52

HardOriginal question

In parallelogram ABCD, $AB = 14$ cm and $AD = 5$ cm. The bisector of $\angle A$ meets side DC at E. What is EC?

2.5 cmCorrect answerYour answer5 cmCorrect answerYour answer7 cmCorrect answerYour answer9 cmCorrect answerYour answer

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CorrectNot quiteAnswer D

Correct answer: D

$AB \parallel DC$ , so $\angle BAE = \angle AED$ (alternate angles).
AE bisects $\angle A$ , so $\angle DAE = \angle BAE = \angle AED$ . Triangle ADE is therefore isosceles with $DE = AD = 5$ cm.
$DC = AB = 14$ cm, so $EC = 14 - 5 = 9$ cm.

Worth noting

5 cm is DE, and 7 cm assumes the bisector hits the midpoint of DC. An angle bisector in a parallelogram cuts off an isosceles triangle.

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