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$$In, triangle, ABC, angle, bisectors, \overline{AD}, and, \overline{BE}, meet, at, point,I. Prove, that, \angle DIB = 90^\circ - \frac{\angle BCA}{2}.$$

Jun 9, 2020

#1
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I don't understand.  Can you explain?

Jun 9, 2020
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I dont understand either that is why i am asking

Guest Jun 9, 2020