解答这个问题！----**Question:** 22. (本小题满分 10 分)
如图, 在 Rt△ABC 中, ∠ACB=90°, AB=5, BC=√21.
(1) 尺规作图: 作∠BAC的角平分线交 BC 于点 P (不写做法, 保留作图痕迹);
(2) 在 (1) 所作图形中, 求△ABP 的面积.
**Handwritten Notes/Calculations:**
AC(=2)
S△ABP = S△ABC - S△ACP
= -
**Diagram Description:**
* **Type:** Geometric figure (Right-angled triangle with construction).
* **Main Elements:**
* **Shape:** A triangle labeled ABC.
* **Vertices:** Labeled points A, B, and C.
* **Angle:** A right angle symbol is shown at vertex C, indicating ∠ACB = 90°.
* **Lines:**
* Line segments forming the sides of the triangle: AC, BC, AB.
* A line segment starting from A and intersecting BC at P, representing the angle bisector of ∠BAC.
* **Points:**
* A, B, C are the vertices.
* P is a point on BC.
* **Construction Marks:** Arcs are drawn from vertex A, intersecting sides AC and AB. Then, arcs of the same radius are drawn from these intersection points, intersecting inside the angle ∠BAC. The line segment AP passes through the intersection of these arcs, indicating the angle bisector construction.
* **Labels:**
* Vertices are labeled A, B, C.
* The intersection point on BC is labeled P.
* The side length BC is labeled with √21 near point P.
**Context Information from Problem Statement:**
* Triangle ABC is a right-angled triangle.
* The right angle is at C (∠ACB = 90°).
* Length of hypotenuse AB = 5.
* Length of side BC = √21.
视频信息
答案文本
视频字幕
Let's analyze this right triangle problem step by step. We have a right triangle ABC with the right angle at C. We know that AB equals 5 and BC equals square root of 21. We need to find the area of triangle ABP where AP is the angle bisector of angle BAC. First, let's find the length of side AC using the Pythagorean theorem.
Now let's apply the Pythagorean theorem to find AC. In a right triangle, the square of the hypotenuse equals the sum of squares of the other two sides. So AC squared plus BC squared equals AB squared. Substituting our known values: AC squared plus 21 equals 25. Therefore AC squared equals 4, which means AC equals 2.
Now let's construct the angle bisector of angle BAC. Using compass and straightedge construction, we first draw an arc from point A that intersects both sides AC and AB. Then from these intersection points, we draw arcs of equal radius that intersect inside the angle. The line from A through this intersection point is the angle bisector AP, which intersects side BC at point P.
Now we apply the Angle Bisector Theorem, which states that the angle bisector divides the opposite side in the ratio of the adjacent sides. So BP over PC equals AB over AC, which is 5 over 2. Since BP plus PC equals BC which is square root of 21, we can solve for the segments. BP equals 5 sevenths times square root of 21, and PC equals 2 sevenths times square root of 21.
Finally, let's calculate the area of triangle ABP. First, we find the area of triangle ABC using the formula one half times base times height. This gives us square root of 21. Since triangles ABP and ABC share the same height from vertex A, the ratio of their areas equals the ratio of their bases. So area of ABP over area of ABC equals BP over BC, which is 5 over 7. Therefore, the area of triangle ABP is 5 square root of 21 over 7.