2
$\begingroup$

A point D is chosen inside an equilateral triangle $ABC$ such that $AD$ = $BD$. A point $E$ outside the triangle is chosen such that $\angle DBE$ = $\angle DBC$ and $BE$ = $AB$. Find the degree measure of angle $\angle DEB$.

My attempt:

I first tried letting $\angle EBD$= x and trying to find the other angles in terms of x, in the hope of getting a congruent triangle. But that didn't go anywhere, or lead to anything useful. It seems that this problem requires some construction to find something equal to $E$ but I can't figure that out

$\endgroup$
  • $\begingroup$ Could you clarify what is angle $E$? Is it $\angle BEA$? $\endgroup$ – Momo May 28 at 13:44
  • $\begingroup$ @Momo Its angle DEB $\endgroup$ – MNIShaurya May 28 at 13:51
  • $\begingroup$ Have you tried drawing it ? I think the answer appears quite naturally, making the proof easier $\endgroup$ – Popyaitte May 28 at 14:19
  • $\begingroup$ @Popyaitte I have. It dosen't seem natural to me, the diagram is really messy-ish. Maybe I'm doing too many unnecessary constructions $\endgroup$ – MNIShaurya May 28 at 14:38
1
$\begingroup$

enter image description here Note that $\angle DCB = \frac12\angle C = 30^\circ$ because of $AD = BD$. Since $\angle EBD = \angle CBD $, $EB = AB = BC$ and $DB=DB$, the triangles $CBD$ and $EBD$ are congruent, which yields $\angle DEB = \angle DCB = 30^\circ$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Construct $DF\perp BC, E\in BC$, and $DG\perp EB, E\in EB$. Then triangles $BFD$ and $BGD$ are congruent, and so are $DGE$ and $DFC$. So $\angle DEB=\angle DCB = 30^\circ$. The construction idea comes from the fact that $D$ is by construction the incenter of the triangle formed by $BE$, $BC$, and $AC$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Maybe you can see this as a proof with almost no words. Just think of reflection in the line through $B$ and $D$:

enter image description here

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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