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

| April 18, 2021 | education

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

Let the angle DAC = X0, then the angle BAC = 2 * X, and since the triangle ABC is isosceles, the angle BCA = BAC = 2 * X0.

The angle ADC and the angle ADB are adjacent angles, then the angle ADC = 180 – 105 = 75.

In triangle ACD, the sum of the interior angles is 180.

DAC + ACD + ADC = X + 2 * X + 75 = 180.

3 * X = 105.

X = 35.

Then the angle BAC = BCA = 2 * 35 = 70.

Angle ABC = 180 – 70 – 70 = 40.

In triangle ABD, the angle ADB is the greatest then side AB> AD.

In triangle ADC, the largest angle is ADC = 75, then the AC side is the largest, AC> AD.

Answer: The angles of the triangle ABC are equal to 70, 40, 70, the segment AD is less than the sides of the triangle ABC.



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