Help!

Hey guys just about a month and a half ago i participated in the national math olympiad in my country organised by a cambridge school in my country.I finished up second and among the problems i didnt solve was also this one.I'm posting it here to see if anyone can help.I would really appreciate it. So,here's the problem.In the figure above $ABC$ is an iscoceles triangle $AB=AC$ . $AD$ is drawn such that $AD=AB=AC$ .If it is known that $CAE=28$ degrees.Find $EBD$ .

#Geometry

Easy Math Editor

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Comments

Because of the given fact, we can draw a circle centred at A through B, C and D. Then using angle at the centre is twice that at the circumference, we can deduce that $\angle EBD = 14°$ . Hope this helps!

Michael Ng - 6 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$