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The second day of the first volume of mathematics supplementary answers
To solve this problem, we need to analyze and reason according to the given information and geometric knowledge.

First of all, according to the description of the topic, we draw a square ABCD, diagonal BD, point P on the ray CD, and connect it with AP. Then, the triangle ADP is translated so that the point D moves to the point C, and a new triangle BCQ is obtained.

Then, the intersection q is that QH is parallel to AD and equal to the length of AD. Because ABCD is a square, AD=BC. At the same time, BD is the diagonal of the square, so BD will divide the angle BDC and the angle BCD equally.

Now, we need to guess which line segments in the graph have a special quantity or positional relationship. According to the tips in the topic, we can list the following relationships:

DA = BQ, because △ADP translates into △BCQ, so their corresponding sides are equal.

AP = AB, because after the translation of △ADP, point P will also move to the corresponding position of point B..

AH = PH, because QH is parallel to AD and QH = AD, so ah is also equal to PH.

HD = HQ, because QH is parallel to AD and QH = AD, HD is also equal to HQ.

Finally, we need to find out how the line segment relationship changes when the point P moves on the line segment CD. Since ABCD is square and AD = BC, when the point P moves on the line segment CD, the length of AP will change, but the length of AB will not change. Therefore, the ratio of AP/AB will change with the position of point P. ..

To sum up, we can draw the conclusion that:

DA = BQ

AP = AB

AH = PH

HD = headquarters

The ratio of AP/AB varies with the position of point P on the CD.

This is the idea and answer to the third question.