Prove that the bisector of the outer angle at the apex of an isosceles triangle opposite to the base is parallel to the rigging?
Let us denote the triangle given by the condition ABC, AB = BC, AC – base, KBC – external angle, BH – bisector of the external angle.
The degree measure of the external angle is equal to the sum of two angles not adjacent to it. In our case, these are two angles at the base, which are equal.
∠ KBC = ∠ BAC + ∠ ACB
The bisector VN divides the KBС angle into two equal angles:
∠ КВС = ∠ КВН + ∠ НВС.
We get that:
∠ АСB = ∠ НВС.
In turn, these two angles are crosswise at the straight lines BH, AC and secant BC.
The equality of the criss-crossing angles gives us the right to assert that the straight lines BN (bisector) and AC (base) are parallel.
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