Question Number 220726 by mr W last updated on 18/May/25
Commented by mr W last updated on 18/May/25
$${an}\:{other}\:{method}\:{for}\:{Q}\mathrm{220630} \\ $$$${without}\:{trigonometry} \\ $$
Commented by mr W last updated on 18/May/25
$$\Delta{BCD}\:{is}\:{equilateral}\:\left({given}\right) \\ $$$$\angle{BAD}=\mathrm{30}°\:\left({given}\right) \\ $$$${make}\:{AE}={AD}\:{and}\:{DE}={DA} \\ $$$$\Rightarrow\Delta{ADE}\:{is}\:{equilateral} \\ $$$$\angle{BAE}=\mathrm{30}°+\mathrm{60}°=\mathrm{90}° \\ $$$$\Rightarrow\Delta{BAE}\:{is}\:{right}\:{angled}\:{triangle} \\ $$$$\angle{BDE}=\angle{BDA}+\mathrm{60}°=\angle{CDA} \\ $$$$\Rightarrow\Delta{BDE}\equiv\Delta{CDA} \\ $$$$\Rightarrow{BE}={AC} \\ $$$${AB}^{\mathrm{2}} +{AE}^{\mathrm{2}} ={BE}^{\mathrm{2}} \\ $$$$\Rightarrow{AB}^{\mathrm{2}} +{AD}^{\mathrm{2}} ={AC}^{\mathrm{2}} \:\checkmark \\ $$
Commented by Mingma last updated on 24/May/25
Prof, can you please explain why angle BDA+60=angleCDA. Also, the congruency of two triangles needs to be supported by explicit conditions like SAS. Thank you
Commented by mr W last updated on 24/May/25
$$\Delta{BCD}\:{is}\:{equilateral}\:\Rightarrow{BD}={CD} \\ $$$$\Delta{ADE}\:{is}\:{equilateral}\:\Rightarrow{DE}={DA} \\ $$$$\angle{BDE}=\angle{BDA}+\angle{ADE}=\angle{BDA}+\mathrm{60}° \\ $$$$\angle{CDA}=\angle{BDA}+\angle{CDB}=\angle{BDA}+\mathrm{60}° \\ $$$$\Rightarrow\angle{BDE}=\angle{CDA} \\ $$$$\Rightarrow\Delta{BDE}\equiv\Delta{CDA}\:\:\left({SAS}\right) \\ $$