Ta có: \(AD = AE + DE\)
Suy ra: \(DE = AD - AE=17 - 8 = 9 (cm)\)
\(\displaystyle {{AB} \over {DE}} = {6 \over 9} = {2 \over 3}\)
\(\displaystyle {{AE} \over {DC}} = {8 \over {12}} = {2 \over 3}\)
\(\Rightarrow \displaystyle {{AB} \over {DE}} ={{AE} \over {DC}} = {2 \over 3}\)
Xét \(∆ ABE\) và \(∆ DEC\) có:
\(\widehat A = \widehat D = 90^\circ \)
\(\displaystyle {{AB} \over {DE}} = {{AE} \over {DC}}= {2 \over 3}\)
\(\Rightarrow ∆ ABE \backsim ∆ DEC \) (c.g.c)
\(\Rightarrow \widehat {ABE} = \widehat {DEC}\) (1)
Xét \(∆ ABE\) có \(\widehat A = 90^\circ\)
\( \Rightarrow \widehat {ABE} + \widehat {AEB} = 90^\circ \) (2)
Từ (1) và (2) suy ra: \( \widehat {DEC} + \widehat {AEB} = 90^\circ \)
Lại có: \(\widehat {AEB} + \widehat {BEC} + \widehat {DEC} = \widehat {AED} \)\(\,= 180^\circ \) (góc bẹt)
\(\Rightarrow \widehat {BEC} = 180^\circ - \left( {\widehat {AEB} + \widehat {DEC}} \right) \)\(\,= 180^\circ - 90^\circ = 90^\circ \).
