2. As shown in Figure 2, in quadrilateral ABCD, AB=10, BC=17, CD=13, DA=20, AC=21. Then BD=
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Solution
2.105. As shown in Figure 6, draw BE⊥AC at point E, DF⊥AC at point F, and draw BG//AC intersecting the extension of DF at point G. Then, quadrilateral BEFG is a rectangle, and △BDG, △ABE, and △CDF are all right triangles. In △ABC, we have p=21(AB+BC+CA)=24.
Thus, BE=21224(24−10)(24−17)(24−21)=8. Similarly, in △ADC, we have DF=12. Therefore, DG=DF+FG=DF+BE=20. Also, AE=AB2−BE2=6, CF=CD2−DF2=5,
Then, BG=EF=AC−AE−CF=10. Thus, BD=DG2+BG2=105.
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