Question

Hi. I have got a really interesting geometry question.

Take any equilateral triangle and pick a random point inside the triangle.Draw from each vertex a line to the random point. Two of the three angles at the random point are known, let’s say angles x and y.If the three line segments from each vertex to the random point were removed out of the original triangle to form a new triangle, what would the new triangle’s angles be?

Source : Oral Entrance Exams.
Department Of Moscow State University.

Can you solve..?
All the best! ♡​

Answer 1

$\huge{\underline{\underline{\mathrm{\red{Question-}}}}}$

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Take any equilateral triangle and pick a random point inside the triangle.Draw from each vertex a line to the random point. Two of the three angles at the random point are known, let’s say angles x and y.If the three line segments from each vertex to the random point were removed out of the original triangle to form a new triangle, what would the new triangle’s angles be?

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Required angles are : $\tt{(x-60°),(y-60°)\:and\:(300°-x-y)}$

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$\huge{\underline{\underline{\mathrm{\red{Explanation-}}}}}$

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Take any equilateral triangle and pick a random point inside the triangle. Now let we take another point opposite to D, such that ∆BED is an equilateral ∆.

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$\therefore$ BD = DE = BE

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$\implies$ $\tt{\angle{DBA}=\angle{EBC}=60°-\angle{DBC}}$

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[ °•° $\sf{\angle{ABC}=\angle{DBE}=60°}$ ]

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Hence, we can say that ∆DBA is congruent to ∆EBC,

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$\implies$ DA = CE [ Corresponding parts of congruent ∆s are equal ]

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$\therefore$ DE = BD

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Now, we get another ∆CDE,

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Let $\angle{ADB}$ = x

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$\angle{BDC}$ = y

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$\large{\boxed{\green{\angle{EDC}=y-60°}}}$

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$\large{\boxed{\green{\angle{DEC}=x-60°}}}$

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$\large{\boxed{\green{\angle{DCE}=300°-x-y}}}$

$\rule{300}1$