Question

[Đỗ Thị Thu Hoài]

Cho tam giác ABC vuông tại A; AB = 15cm, AC = 20cm, đường cao AH
a) Chứng minh: tam giác HBA đồng dạng với tam giác ABC
b) Tính BC, AH
c) Gọi E là điểm đối xứng vói B qua H.vẽ hình bình hành ADCE.tứ giácABCE LÀ hình gì? tại sao            

d, tính AE

Answer 1

[Lê Song Phương]

c) This makes no sense. What is quadrilateral ABCE while B, E, C are collinear? May be you mean ABCD but you make some mistake when typing.

ADCE is a parallelogram, therefore, \(CE//AD\) or \(BC//AD\)

Thus, we can easily prove that ABCD is a trapezoid.

Consider the triangle ABE, its height AH is also a median. This means ABE is an isosceles triangle, whose bisector is also AH. Thus, \(\widehat{BAH}=\widehat{EAH}\)

Clearly, we get \(\widehat{BAH}=\widehat{ECA}\left(=90^o-\widehat{B}\right)\), so, \(\widehat{EAH}=\widehat{ECA}\left(=\widehat{BAH}\right)\)

On the other hand, \(EC//AD\Rightarrow\widehat{ECA}=\widehat{CAD}\) (2 staggered angles)

Thus, \(\widehat{EAH}=\widehat{CAD}\left(=\widehat{ECA}\right)\) or \(\widehat{EAH}+\widehat{EAC}=\widehat{CAD}+\widehat{EAC}\) or \(\widehat{CAH}=\widehat{DAE}\)

ADCE is a parallelogram, therefore, \(\widehat{DAE}=\widehat{DCE}\), so, \(\widehat{CAH}=\widehat{DCE}\left(=\widehat{DAE}\right)\) or \(\widehat{CAH}=\widehat{DCB}\)

We also have \(\widehat{CAH}=\widehat{B}\left(=90^o-\widehat{ACB}\right)\), so \(\widehat{B}=\widehat{DCB}\left(=\widehat{CAH}\right)\)

Consider the trapezoid ABCD (AD//BC), it has \(\widehat{B}=\widehat{DCB}\), therefore, ABCD is an isosceles trapezoid.

d) It's pretty easy! Since ABE is an isosceles triangle, \(AE=AB\). Guess what? We've already had \(AB=15cm\). So, simply, we get \(AE=15cm\) !!!