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1. Identify vertex angle (ΔCDE)
In an isosceles triangle, the vertex angle is between the two congruent sides. For $\Delta CDE$ with $\overline{CE} \cong \overline{CD}$, the vertex angle is $\angle C$.
2. Identify vertex angle (ΔEBC)
For $\Delta EBC$ with $\overline{BE} \cong \overline{CB}$, the vertex angle is between these sides: $\angle B$.
3. Identify congruent angles (ΔAEB)
In $\Delta AEB$, $\overline{AE} \cong \overline{BE}$, so the base angles opposite these sides are congruent: $\angle A \cong \angle B$.
4. Identify congruent sides (ΔBCE)
In $\Delta BCE$, $\angle BCE \cong \angle BEC$, so the sides opposite these angles are congruent: $\overline{BE} \cong \overline{CB}$.
5. Identify base of ΔAEB
If $\angle E$ is the vertex angle of isosceles $\Delta AEB$, the base is the side opposite the vertex angle: $\overline{AB}$.
6. Solve for x (ΔDEF)
ΔDEF is equilateral (all sides =5), so all angles =60°. Set $3x = 60$.
$x = \frac{60}{3} = 20$
7. Solve for x (ΔJKL)
ΔJKL is isosceles ($\overline{JK} \cong \overline{JL}$), so opposite sides are equal. Set $5x + 5 = 35$.
$5x = 35 - 5 = 30$
$x = \frac{30}{5} = 6$
8. Solve for x (ΔABC)
ΔABC is isosceles ($\overline{AB} \cong \overline{AC}$), so base angles are equal. Sum of angles =180°.
$9x + 9x + 72 = 180$
$18x = 180 - 72 = 108$
$x = \frac{108}{18} = 6$
9. Solve for x,y (Isosceles Triangle)
The triangle is isosceles with $x = y$. Sum of angles =180°.
$x + y + 102 = 180$
$2x = 180 - 102 = 78$
$x = 39$, $y = 39$
10. Solve for x,y (Isosceles Triangle)
The triangle is isosceles, so $(x+7) = 55$.
$x = 55 - 7 = 48$
Sum of angles =180°: $y + 55 + 55 = 180$
$y = 180 - 110 = 70$
11. Solve for x,y (Right Isosceles Triangle)
The triangle is right isosceles, so $x = 9y$, and $x + 9y + 90 = 180$.
Substitute $x=9y$: $9y + 9y = 90$
$18y = 90$
$y = 5$, $x = 9(5) = 45$
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- $\angle C$
- $\angle B$
- $\angle A \cong \angle B$
- $\overline{BE} \cong \overline{CB}$
- $\overline{AB}$
- $x = 20$
- $x = 6$
- $x = 6$
- $x = 39$, $y = 39$
- $x = 48$, $y = 70$
- $x = 45$, $y = 5$