In the diagram below, AD is perpendicular to AC and $ ∠BAD = ∠DAE = 12^\circ$. If $AB + AE = BC$, find $∠ABC$. Blockquote

The above is the diagram.

I came across this question in a Math Olympiad Competition and I am not sure how to solve it.


closed as off-topic by heropup, user122283, colormegone, Davide Giraudo, user7530 Jun 14 '14 at 16:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Community, colormegone, Davide Giraudo, user7530
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I do not actually know how to do the question on first sight. I am only able to find out the angles of $∠BAD$ , $∠DAE$ and $∠EAC$ but after that, I am not sure how to solve it. $\endgroup$ – snivysteel Jun 15 '14 at 2:33


Take $F$ on $\overrightarrow{BA}$ beyond $A$ such that $BF=BC$
$AE=AF$ and $\triangle AEF$ is isosceles.
$2\angle BAD=\angle BAE=\angle AEF+\angle AFE=2\angle AFE$
Thus $AD$ is parallel to $EF$.
$AD\bot AC$ and then $AC\bot EF$, which tell us $\square AECF$ is kite.
$\angle EFC=90^{\circ}-\angle FCA$, $\angle AFC=\angle AFE+\angle CFE=102^{\circ}-\angle FCA$
$\angle AFC=\angle ECF=2\angle FCA$, since $\triangle BCF$ is isosceles.
Hence, $\angle FCA=34^{\circ}$ and $\angle ABC=44^{\circ}$


Not the answer you're looking for? Browse other questions tagged or ask your own question.