In the given figure, if KI = IT and EK = ET, then ∠TEI = _______.

Calculation

In \(\Delta \)KEI and \(\Delta \)TEI

⇒ KI = IT (Given)

⇒ EK = ET (Given)

⇒ EI = EI (Common)

\(\Delta \)KEI is congruent to \(\Delta \)TEI  (by SSS)

⇒ \(\angle\)KEI = \(\angle\)TEI (By CPCT)

Now,

⇒ \(\angle\)KET + \(\angle\)KEI + \(\angle\)TEI = \({360}^\circ\)

⇒ \({150}^\circ\) + 2 x \(\angle\)TEI = \({360}^\circ\)

⇒ 2\(\angle\)TEI = \({360}^\circ\) - \({150}^\circ\)

⇒ \(\angle\)TEI = \(\frac{210}{2}\) = \({105}^\circ\)

Therefore, the answer is \({105}^\circ\)