Cho hình vẽ : Biết BAC + ACD =180 độ ; ADC=40 độ; BAC=130 độ Chứng minh AD vuông góc AC Pờ li làm hộ đi November 14, 2020 by Cherry Cho hình vẽ : Biết BAC + ACD =180 độ ; ADC=40 độ; BAC=130 độ Chứng minh AD vuông góc AC Pờ li làm hộ đi
$\widehat{BAC}+\widehat{ACD}=180^\circ$ mà $\widehat{BAC}=130^\circ$ $→\widehat{ACD}=180^\circ-130^\circ=50^\circ$ Xét $ΔACD$: $\widehat{ACD}+\widehat{ADC}+\widehat{CAD}=180^\circ$ (tổng 3 góc trong 1 Δ) $→\widehat{CAD}=180^\circ-\widehat{ACD}-\widehat{ADC}=180^\circ-40^\circ-50^\circ=90^\circ$ $→AC⊥AD$ (ĐPCM) Reply
$\widehat{BAC}+\widehat{ACD}=180^o$ $\Leftrightarrow \widehat{ACD}=180^o-\widehat{BAC}=50^o$ $\Delta ACD$ có: $\widehat{CAD}+\widehat{ACD}+\widehat{ADC}=180^o$ $\Leftrightarrow \widehat{CAD}=180^o-50^o-40^o=90^o$ $\to AC\bot AD$ Reply
$\widehat{BAC}+\widehat{ACD}=180^\circ$
mà $\widehat{BAC}=130^\circ$
$→\widehat{ACD}=180^\circ-130^\circ=50^\circ$
Xét $ΔACD$:
$\widehat{ACD}+\widehat{ADC}+\widehat{CAD}=180^\circ$ (tổng 3 góc trong 1 Δ)
$→\widehat{CAD}=180^\circ-\widehat{ACD}-\widehat{ADC}=180^\circ-40^\circ-50^\circ=90^\circ$
$→AC⊥AD$ (ĐPCM)
$\widehat{BAC}+\widehat{ACD}=180^o$
$\Leftrightarrow \widehat{ACD}=180^o-\widehat{BAC}=50^o$
$\Delta ACD$ có:
$\widehat{CAD}+\widehat{ACD}+\widehat{ADC}=180^o$
$\Leftrightarrow \widehat{CAD}=180^o-50^o-40^o=90^o$
$\to AC\bot AD$