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Bài 1:
b)
chứng minh EDCB là tgnt => góc AED = góc ACB
từ đó, chứng minh tam giác AED đồng dạng ACB (gg)
=> DE / BC = AD / AB
tam giác ADB vuông tại A => AD / AB = cotg A = cotg 45 = 1
c)
kẻ tiếp tuyến tại Ax của (O) (Ax thuộc nửa mp bờ AC chứa B)
góc xAB = ACB = AED
=> DE // Ax
Mà Ax vuông góc với OA nên OA vuông góc với DE. (đpcm)
Gọi GE,FD cắt đường tròn (O) lần thứ hai tại H,I.
Ta thấy F nằm trên trung trực BD => ΔΔBDF cân tại F. Mà ΔΔBDF ~ ΔΔIDA (g.g) nên ΔΔIDA cân tại A
Hay AI = AD. Tương tự ta có AH = AE. Do AD = AE nên AH = AD = AE = AI => A cách đều 4 điểm H,D,E,I
=> Tứ giác DEIH nội tiếp. Vậy thì ^DEH = ^DIH = ^HIF = ^HGF => DE // FG (2 góc đồng vị bằng nhau) (đpcm).
Draw your own figure. Now I'll give the solution:
a) Because BD and CE are the heights of the triangle ABC \(\left(D\in AC;E\in AB\right)\), we have \(\widehat{BEC}=\widehat{BDC}=90^o\)
Consider the quadrilateral BCDE, it has 2 adjacent vertices D, E which both look at the edge BC by a right angle. Thus, BCDE is an inscribed quadrilateral. And that's what we must prove!
b) We can easily have \(\widehat{EBD}=\widehat{ECD}\) due to the inscribed quadrilateral BCDE or \(\widehat{ABD}=\widehat{HCD}\)
Consider the 2 triangles DAB and DHC, which are both right at D, have \(\widehat{ABD}=\widehat{HCD}\). Therefore, \(\Delta DAB~\Delta DHC\left(a.a\right)\). This means \(\dfrac{DA}{DH}=\dfrac{DB}{DC}\) or \(DA.DC=DH.DB\) and again, that's what we must prove!
c) Draw the tangent Ax of (O). We have \(Ax\perp OA\) (at A)
Consider the circle (O), it has \(\widehat{BAx}\) is an angle that is formed by the tangent line Ax and the chord AB. Also, \(\widehat{ACB}\) is the inscribed angle intercept the arc AB. Therefore, \(\widehat{BAx}=\widehat{ACB}\)
On the other hand, \(\widehat{ACB}=\widehat{AED}\) due to the inscribed quadrilateral BCDE. Thus, we must have \(\widehat{BAx}=\widehat{AED}\), which means \(Ax//DE\) (because the 2 staggered angles are equal). We have already prove \(Ax\perp OA\), so, \(OA\perp DE\)
Consider the circle (H), we have \(HE\perp AM\) at E while AM is a chord of (H). Therefore, E is the midpoint of AM.
Similarly, D is the midpoint of AN.
Consider the triangle AMN, it has D, E, consecutively, are the midpoint of AN, AM. Thus, DE must be the average line of the triangle AMN. This means \(DE//MN\)
Guess what? We've already had \(OA\perp DE\), so, \(OA\perp MN\), and that's what we must prove!
d) Sorry, I haven't had the solution for this yet.