Track / Stage 4 / 298 of 340 #558 of 2000

Problem 558

AMC 12 late, AIME early

Geometry Difficulty 5.0 Find the answer

4. As shown in Figure 4, in $\triangle A B C$ , $A B=A C$ , $\angle B=40^{\circ}$ , $B D$ is the angle bisector of $\angle B$ , and $B D$ is extended to $E$ such that $D E=A D$ . Then the degree measure of $\angle E C A$ is

Your answer

Official solution

4. $40^{\circ}$ .

As shown in Figure 7, on $BC$ , take $BF=AB$ , connect $DF$ , then $\triangle ABD \cong \triangle FBD$ .
$\therefore DF=DA=DE.$

From $AC=AB$ , we know $\angle ACB=40^{\circ}$ .
$\begin{aligned} \angle DFC & =180^{\circ}-\angle DFB \\ & =180^{\circ}-80^{\circ}=100^{\circ}, \end{aligned}$
$\therefore \angle FDC=60^{\circ}.$

Also, $\angle BDF=180^{\circ}-20^{\circ}-100^{\circ}=60^{\circ}$ , so $\angle CDE=60^{\circ}$ ,
$\therefore \triangle DCF \cong \triangle DCE$ .
Thus, $\angle DCE=\angle DCF=40^{\circ}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.