Question by Summer Clouds - Discuss with MathYouLike

Summer Clouds moderators

26/06/2017 at 09:14

In quadrilateral ABCD, AB // CD, \(\angle ABC=125^o\) and \(\angle ADC=75^o\). Show that \(CD=AB+AD\).

Suggest: Draw line AE // BC.

List of answers

Trần Nhật Dương 26/06/2017 at 15:32

Draw AE // BC.

The quadrilateral ABCE has AB // CE ; AE // BC,so it's a parallelogram

$\Rightarrow A B = C E; \hat{E_{1}} = \hat{B} = 125^{0} \Rightarrow \hat{E_{2}} = 180^{0} - \hat{E_{1}} = 55^{0}$

$\hat{E_{1}}$ is the exterior angle of $Δ A D E$ $\Rightarrow \hat{A_{1}} + \hat{D} = \hat{E_{1}}$

$\Rightarrow \hat{A_{1}} = 125^{0} - 70^{0} = 55^{0}$

$Δ A D E$ has $\hat{A_{1}} = \hat{E_{2}} = 55^{0}$ ,so $Δ A D E$ isosceles at D

$\Rightarrow A D = D E \Rightarrow C D = C E + D E = A B + A D$

P/S : The question must be changed into $\hat{D} = 70^{0}$ or $\hat{B} = {127.5}^{0}$
Phan Thanh Tinh Coordinator 26/06/2017 at 10:57

Draw AE // BC.

The quadrilateral ABCE has AB // CE ; AE // BC,so it's a parallelogram

\(\Rightarrow AB=CE;\widehat{E_1}=\widehat{B}=125^0\Rightarrow\widehat{E_2}=180^0-\widehat{E_1}=55^0\)

\(\widehat{E_1}\)is the exterior angle of \(\Delta ADE\) \(\Rightarrow\widehat{A_1}+\widehat{D}=\widehat{E_1}\)

\(\Rightarrow\widehat{A_1}=125^0-70^0=55^0\)

\(\Delta ADE\) has \(\widehat{A_1}=\widehat{E_2}=55^0\),so \(\Delta ADE\) isosceles at D

\(\Rightarrow AD=DE\Rightarrow CD=CE+DE=AB+AD\)

P/S : The question must be changed into \(\widehat{D}=70^0\)or \(\widehat{B}=127.5^0\)

Weekly ranking