In an isosceles triangle ABC, the base of the AC is larger than the lateral side. The bisector AD forms angles

| September 4, 2021 | education

In an isosceles triangle ABC, the base of the AC is larger than the lateral side. The bisector AD forms angles with the side BC, one of which is 105 degrees. a) Find the angles of the triangle ABC. b) Compare the segment AD with the sides of the triangle ABC.

Let the angle CAD be equal to X0.

Then the angle BAC = 2 * X, since AD is the bisector of the angle BAC.

The ABC triangle is isosceles, then the BCA angle = BAC = 2 * X0.

In triangle ACD, the sum of the angles (X + 2 * X + 105) = 180.

3 * X = 180 – 105 = 75.

X = 75/3 = 25.

Then the angle BAC = BCA = 2 * 25 = 50, angle ABC = (180 – 50 – 50) = 80.

Determine the angles of the triangle ABD, the angle ADB = 180 – 105 = 75.

Angle ABD = 80.

Angle ABD> ADB, therefore AD> AB.

In the triangle ADC, the angle ADC is the greatest, then AD <AC.

Answer: The angles of the triangle ABC are equal to 50, 80, 50. The median AD is greater than AB and BC, but less than AC.



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