Question

Около четырёхугольника ABCD описана окружность, причём AD – диаметр окружности. Зная, что угол ABC = 112°, угол BCD = 128°, найдите: а) угол BAD б) угол CAD в) угол BDA И, если не сложно, можно рисунок, пожалуйста?очень нужна помощь, заранее спасибо

Answer 1

Ответ с решением и рисунком на фото

Answer 2

Given Information

We are given a quadrilateral ABCD with a circumcircle, where AD is the diameter of the circle. We also know that angle ABC is 112° and angle BCD is 128°.

Solution

To find the angles BAD, CAD, and BDA, we can use the properties of angles in a cyclic quadrilateral.

Angle BAD

Angle BAD is an inscribed angle that intercepts the arc BC. According to the inscribed angle theorem, the measure of an inscribed angle is half the measure of its intercepted arc.

Since AD is the diameter of the circle, the intercepted arc BC is a semicircle. The measure of a semicircle is 180°. Therefore, angle BAD is half of 180°, which is 90°.

Angle CAD

Angle CAD is also an inscribed angle that intercepts the arc BD. Using the same reasoning as above, we can determine that angle CAD is also 90°.

Angle BDA

Angle BDA is an exterior angle of triangle ABD. The sum of the measures of the exterior angles of a triangle is always 360°.

To find angle BDA, we can subtract the sum of angles BAD and BCD from 360°.

Angle BDA = 360° - (angle BAD + angle BCD)

Substituting the given values, we have:

Angle BDA = 360° - (90° + 128°) = 360° - 218° = 142°

Therefore, the angles are: a) Angle BAD = 90° b) Angle CAD = 90° c) Angle BDA = 142°

Here is a diagram to help visualize the quadrilateral ABCD with the given angles:

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