Câu hỏi của minh anh

Question

) Cho tam giác ABC có ba góc nhọn và AB < BC < AC nội tiếp đường tròn (O). Kẻ các đường cao BD, CE cắt nhau tại H (D ∈ AC, E ∈ AB).

1)    Chứng minh BCDE là tứ giác nội tiếp.

2)    Chứng minh DA.D...

Lê Song Phương · Accepted Answer

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.

Câu trả lời: 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.

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