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**Soroban** Hello, yehoram!

$\displaystyle \angle ADE = 15^o \quad\Rightarrow\quad \angle ADC = 75^o$

$\displaystyle \text{Since }\Delta CAD\text{ is isosceles: }\:\angle DAC = 75^o$

$\displaystyle \text{In }\Delta CAD,\;\angle ACD = 30^o\quad\Rightarrow\quad \angle ACB = 60^o$

$\displaystyle \text{Since }\angle AED = 15^o\text{, we can show that: }\:\angle ABC = 60^o$

. . $\displaystyle \text{Hence: }\:\angle BAC = 60^o$

$\displaystyle \text{Therefore, }\Delta ABC\text{ is equilateral.}$