In an isosceles triangle ABC, the bisector BN of the outer angle at apex B is drawn.
| September 21, 2021 | education
In an isosceles triangle ABC, the bisector BN of the outer angle at apex B is drawn. Determine the angle HBN that the bisector BN makes with the height BH drawn to the base of AC.
Let the value of the angle ABC = 2 * X0, then the height BH, which is also the bisector of the angle, divides it in half, then the angle СВН = 2 * X / 2 = X0.
The angle ABC and CBD are adjacent angles, then the angle CBD = (180 – 2 * X) 0.
The segment BN is the bisector of the CBD angle, then the angle CBN = CBD / 2 – (180 – 2 * X) / 2 = (90 – X) 0.
Angle HBN = CBN + СBН = (90 – X + X) = 90.
Answer: The HBN angle is 90.
Answer: The length of the side is 4 * √3 cm, the height is 2 * √3 cm.
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