Ta có
$\dfrac{AH}{DC} = \dfrac{DB}{DC} = \sin \widehat{DCB}$
Do AD//BC nên $\widehat{DCB} = \widehat{ADC}$ nên ta có
$\dfrac{BH}{DE} = \dfrac{AD}{DE} = \cos \widehat{ADC} = \cos \widehat{DCB}$
Theo tchat lượng giác ta có
$\sin^2 \widehat{DCB} + \cos^2 \widehat{DCB} = 1$
$<-> \dfrac{DB^2}{DC^2} + \dfrac{AD^2}{DE^2} = 1$
$<-> \dfrac{AH^2}{DC^2} + \dfrac{BH^2}{DE^2} = 1$ (dpcm)
