2. The length of diagonal ACBD in diamond ABCD is 8 and 6 respectively. Move BD to DC direction, so that point D coincides with point C, and point B falls on point E on AB extension line, and find the area of △BEC.
3. When △AB, AC, ∠ B = 90, E and F are on AB and AC respectively, and are folded in half along EF, so that point A falls on point D on BC, and FD⊥BC. Try to judge the shape of quadrilateral AEDF and prove your conclusion.
4. Try to judge the shapes of quadrilateral E, F, G and H, and explain the reasons. The parallel lines of the four vertices of rectangular ABCD are diagonal lines, and the intersections are E, F, G and H respectively.
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I don't need to write in a standard format. You can convert it yourself.
1, AD is parallel to A 1D 1, so the angle DAC is equal to d1a1. Similarly, DCA = ∠ D 1C 1, and
Because ∠ FA 1C = ∠ DAC = ∠ EA 1C, ∠ FCA = ∠ ECA, A 1C = A 1C, triangle fa/kloc.
2, ∫AE is parallel to CD and BD is parallel to CE, so the quadrilateral BECD is a parallelogram, so the triangle area is equal to half of the parallelogram area, equal to the triangle ABD area, and equal to 6*(8/2)/2= 12.
3. Fold in half, so ∠DAF=∠ADF, ∠AE=ED=∠EDA, because ∠EDA+∠ADF+∠EDB=90 degrees, ∠EDA+∠ADF =∞.
4. Because they are all parallel, they are parallelograms. Let the central focus be O, the quadrilateral OAEB be a parallelogram, and EA=OB. Similarly, AH=OD, then EH equals BD. You can also get AC=HG, because AC=BD, so the quadrilateral is a diamond.
I'll just write a general outline. I am exhausted.