А.с және б.д нүктелерінде көлденең қабырғалары араласында көтерілген төбесі бар түзулермен сәйкесінше қиылысқан

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15

Шахс болыпшылдық - One could consider that triangle AСВ is a right-angled triangle, where AB and BC are the legs, and AC is the hypotenuse. The given problem states that a circle with a diameter lying on segment AB is tangent to the hypotenuse AC, and a circle with a diameter lying on segment BC is tangent to the hypotenuse AC as well. We need to prove that the triangles АСУ and CДС are similar, where point D is the point of tangency between the circles and the hypotenuse.

Ілгері 1 - First, let"s connect the points of tangency D and E, where E is the point of tangency of the circle with diameter BC and the hypotenuse AC.

Ілгері 2 - Since the points of tangency divide the hypotenuse AC into segments, we can consider the lengths of these segments. Let CE = x and DA = y.

Ілгері 3 - By the properties of tangents, we know that the tangent segment from an exterior point to a circle is equal in length to the tangent segment from the same point to the other circle. Therefore, DE = x and AE = y.

Ілгері 4 - Since triangles AСU and CDS are right-angled triangles, we can use the Pythagorean theorem to express the lengths of their sides. Starting with triangle AСU, we have AC^2 = AU^2 + UC^2.

Ілгері 5 - Substituting the lengths, we have (AC/2 + y)^2 = (AC/2 + x)^2 + y^2. Expanding this equation, we get AC^2/4 + ACy + y^2 = AC^2/4 + ACx + x^2 + y^2.

Ілгері 6 - Simplifying the equation, we subtract AC^2/4 from both sides and combine like terms, giving ACy = ACx + x^2.

Ілгері 7 - Now, let"s look at triangle CDS. Using the same logic as before, we have AC^2 = CD^2 + DS^2.

Ілгері 8 - Substituting the lengths, we have (AC/2 + x)^2 = (AC/2 + y)^2 + x^2. Expanding this equation, we get AC^2/4 + ACx + x^2 = AC^2/4 + ACy + y^2 + x^2.

Ілгері 9 - Simplifying the equation, we subtract AC^2/4 from both sides and combine like terms, giving ACx = ACy + y^2.

Ілгері 10 - Comparing equations (6) and (9), we can see that ACy = ACx + x^2 is equivalent to ACx = ACy + y^2. This means that x^2 = y^2, and since lengths cannot be negative, x = y.

Ілгері 11 - Now that we know x = y, we can conclude that DE = x = y = AE. This shows that triangles ASC and SEC are congruent by the Side-Angle-Side (SAS) criterion.

Ілгері 12 - Congruent triangles have corresponding angles that are equal. Therefore, we can conclude that angle AUS = angle CSD and angle ASC = angle SEC.

Ілгері 13 - We can now consider triangles AUS and CDS. The angles ASC and SEC are congruent, and since angles USА and SCD are both right angles, we have two pairs of corresponding angles that are equal.

Ілгері 14 - By the Angle-Angle (AA) criterion, we can conclude that triangles AUS and CDS are similar.

Ілгері 15 - Thus, we have proven that triangles АСУ and CДС are similar, as desired.