A problem by Yahia El Haw

Level pending

Show that in a convex quadrilateral the bisector of two consecutive angles forms an angle whose measure is equal to half the sum of the measures of the other two angles.

m ( C ^ ) + m ( D ^ ) 2 \frac { m\left( \hat { C } \right) +m\left( \hat { D } \right) }{ 2 } m ( C ^ ) − m ( D ^ ) 2 \frac { m\left( \hat { C } \right) -m\left( \hat { D } \right) }{ 2 } m ( D ^ ) − m ( C ^ ) 2 \frac { m\left( \hat { D } \right) -m\left( \hat { C } \right) }{ 2 } m ( C ^ ) + m ( B ^ ) 2 \frac { m\left( \hat { C } \right) +m\left( \hat { B } \right) }{ 2 }

Yahia El Haw by Yahia El Haw , 18, Egypt

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1 solution

Yahia El Haw
Dec 1, 2016

m ( A E B ^ ) = m ( D ^ ) + m ( C ^ ) 2 m\left( \hat { AEB } \right) = \frac { m\left( \hat { D } \right) +m\left( \hat { C } \right) }{ 2 }

m ( A ^ ) + m ( B ^ ) + m ( C ^ ) + m ( D ^ ) = 360 ∘ m\left( \hat { A } \right) +m\left( \hat { B } \right) +m\left( \hat { C } \right) +m\left( \hat { D } \right) ={ 360 }^{ \circ }

m ( A ^ ) + m ( B ^ ) 2 = 180 ∘ − m ( C ^ ) + m ( D ^ ) 2 \frac { m\left( \hat { A } \right) +m\left( \hat { B } \right) }{ 2 } ={ 180 }^{ \circ }-\frac { m\left( \hat { C } \right) +m\left( \hat { D } \right) }{ 2 }

m ( A E B ^ ) = 180 ∘ − m ( A ^ ) 2 − m ( B ^ ) 2 m\left( \hat { AEB } \right) ={ 180 }^{ \circ }-\frac { m\left( \hat { A } \right) }{ 2 } -\frac { m\left( \hat { B } \right) }{ 2 }

= 180 ∘ − 180 ∘ + m ( C ^ ) + m ( D ^ ) 2 = m ( C ^ ) + m ( D ^ ) 2 ={ 180 }^{ \circ }-{ 180 }^{ \circ }+\frac { m\left( \hat { C } \right) +m\left( \hat { D } \right) }{ 2 } =\frac { m\left( \hat { C } \right) +m\left( \hat { D } \right) }{ 2 }

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