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How to prove the judgment of diamonds
Math teaching case in grade three: Section 1.33

Title: Proof of Diamond Property Theorem

Teaching goal: to master the nature judgment of diamonds, so that students can flexibly use diamond knowledge to solve related problems and improve their ability.

Through the definition and nature of rectangle and diamond, the easily confused knowledge points are clearly separated, so as to cultivate students' correct views.

Teaching emphasis: proof of diamond property theorem

Teaching difficulties: the mutual transformation between life mathematics and pure mathematics by applying the theorem of nature.

teaching process

Knowledge review:

1._ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ (Page 8 12)

The diamond is also a special parallelogram, because it has the characteristics of parallelogram ① _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

2. There is a special area formula: _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.

3. As shown in the figure, the high bisector BC of the diamond-shaped ABCD is ∠ BAE = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.

Newly granted content:

Prove that the four sides of a diamond are equal.

1. parallelogram ABCD is known, AB=AD, verify.

① B=BC=CD=DA

2. Given the rhombus ABCD, the diagonals intersect at O, can it be proved that the diagonals are perpendicular to each other?

3. The diamond ABCD whose diagonal intersects with O proves that AC and BD are equally divided.

Example 3: As shown in Figure 3, the movable coat rack consists of three congruent diamonds. Vertex A, E, F, C, G and H are two rows of hooks, and the distance between the hooks can be changed as needed (for example, AC two points can move up and down freely). If the side length of the diamond is 13 cm, and the distance between two rows of hooks should be 24 cm, fixed at points B and M, then B.

Tip: This is an example in life. How to solve such a problem? First of all, everyone should turn the mathematical problems in life into pure mathematics in our textbooks to answer.

Homework done in the classroom

(1) It is known that the area of the diamond is 30 square centimeters. If the diagonal length is 12 centimeters, don't take the diagonal length as _ _ _ _ _ centimeters.

② In the rhombic ABCD, ∠ABC=60 degrees, BD= 10, The AC length of the diagonal is _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

As shown in the figure, AD is △ABC, DE//AC, intersects with AB at point E, DF//AB, and intersects with AC at point F. It is proved that AD⊥EF.

Class Summary: Mathematics problems are living and mathematics theories are living.

There is no "difficult" mathematics, only difficult psychology. Adjust your mentality. As long as I do my best, I will be worthy of myself and my family.

Although you can't reach the sky in one step, you can make progress every day, even a little bit, which is very fulfilling.

It is known that in diamond-shaped ABCD, e and f are points on the sides of BC and CD respectively, and CE=CF,

① Verification of△ ①△ABE?△ADF

② Let CG//EA pass through point C and AD pass through point G, and if ∠BAE=30 degrees and BCD = 130 degrees, find ∠AHC.