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Question Number 225885 by Mingma last updated on 15/Nov/25
Commented by Mingma last updated on 15/Nov/25
Can you show workings, Prof?
Commented by fantastic2 last updated on 15/Nov/25
∠DCH+∠DBH=45^0
$$\angle{DCH}+\angle{DBH}=\mathrm{45}^{\mathrm{0}} \\ $$
Answered by A5T last updated on 16/Nov/25
Let AB=AD=x and ∠EBD=∠DBC=y° BD=(√(x^2 +x^2 ))=x(√2) ∠ACB=45°−y; ∠DEB=135°−y ∠BEH=90°−2y ⇒ ∠HED=45°+y [△BDC]((sin(45°−y))/(BD))=((sin135°)/(BC))⇒BC=(x/(sin(45°−y))) [△BDC]((sin135°)/(BC))=((siny)/(CD))⇒CD=((x(√2)sin(y))/(sin(45°−y))) [△BED]((sin45°)/(BE))=((sin(135°−y))/(BD))⇒BE=(x/(sin(135°−y))) [△BED]((siny)/(DE))=((sin45°)/(BE))⇒DE=((x(√2)siny)/(sin(135°−y))) [△BEH]((sin2y)/(EH))=((sin90)/(BE))⇒EH=((xsin(2y))/(sin(135°−y))) [CEH]((sin90°)/(CE))=((sin(45°−y))/(EH))⇒CE=((xsin(2y))/(sin(135°−y)sin(45°−y))) ((sin90°)/(CE))=((sin45°+y)/(CH))⇒CH=(((√2)xsin(2y))/((cosy−siny))) ((CH)/(CD))=(√2)cosy; ((HE)/(ED))=(√2)cosy ((CH)/(CD))=((HE)/(ED)) ⇒ HD bisects ∠CHE ⇒ ∠CHD=((90°)/2) ⇒∠CHD=45°
$$\mathrm{Let}\:\mathrm{AB}=\mathrm{AD}=\mathrm{x}\:\mathrm{and}\:\angle\mathrm{EBD}=\angle\mathrm{DBC}=\mathrm{y}° \\ $$$$\mathrm{BD}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }=\mathrm{x}\sqrt{\mathrm{2}} \\ $$$$\angle\mathrm{ACB}=\mathrm{45}°−\mathrm{y};\:\angle\mathrm{DEB}=\mathrm{135}°−\mathrm{y} \\ $$$$\angle\mathrm{BEH}=\mathrm{90}°−\mathrm{2y}\:\Rightarrow\:\angle\mathrm{HED}=\mathrm{45}°+\mathrm{y} \\ $$$$\left[\bigtriangleup\mathrm{BDC}\right]\frac{\mathrm{sin}\left(\mathrm{45}°−\mathrm{y}\right)}{\mathrm{BD}}=\frac{\mathrm{sin135}°}{\mathrm{BC}}\Rightarrow\mathrm{BC}=\frac{\mathrm{x}}{\mathrm{sin}\left(\mathrm{45}°−\mathrm{y}\right)} \\ $$$$\left[\bigtriangleup\mathrm{BDC}\right]\frac{\mathrm{sin135}°}{\mathrm{BC}}=\frac{\mathrm{siny}}{\mathrm{CD}}\Rightarrow\mathrm{CD}=\frac{\mathrm{x}\sqrt{\mathrm{2}}\mathrm{sin}\left(\mathrm{y}\right)}{\mathrm{sin}\left(\mathrm{45}°−\mathrm{y}\right)} \\ $$$$\left[\bigtriangleup\mathrm{BED}\right]\frac{\mathrm{sin45}°}{\mathrm{BE}}=\frac{\mathrm{sin}\left(\mathrm{135}°−\mathrm{y}\right)}{\mathrm{BD}}\Rightarrow\mathrm{BE}=\frac{\mathrm{x}}{\mathrm{sin}\left(\mathrm{135}°−\mathrm{y}\right)} \\ $$$$\left[\bigtriangleup\mathrm{BED}\right]\frac{\mathrm{siny}}{\mathrm{DE}}=\frac{\mathrm{sin45}°}{\mathrm{BE}}\Rightarrow\mathrm{DE}=\frac{\mathrm{x}\sqrt{\mathrm{2}}\mathrm{siny}}{\mathrm{sin}\left(\mathrm{135}°−\mathrm{y}\right)} \\ $$$$\left[\bigtriangleup\mathrm{BEH}\right]\frac{\mathrm{sin2y}}{\mathrm{EH}}=\frac{\mathrm{sin90}}{\mathrm{BE}}\Rightarrow\mathrm{EH}=\frac{\mathrm{xsin}\left(\mathrm{2y}\right)}{\mathrm{sin}\left(\mathrm{135}°−\mathrm{y}\right)} \\ $$$$\left[\mathrm{CEH}\right]\frac{\mathrm{sin90}°}{\mathrm{CE}}=\frac{\mathrm{sin}\left(\mathrm{45}°−\mathrm{y}\right)}{\mathrm{EH}}\Rightarrow\mathrm{CE}=\frac{\mathrm{xsin}\left(\mathrm{2y}\right)}{\mathrm{sin}\left(\mathrm{135}°−\mathrm{y}\right)\mathrm{sin}\left(\mathrm{45}°−\mathrm{y}\right)} \\ $$$$\frac{\mathrm{sin90}°}{\mathrm{CE}}=\frac{\mathrm{sin45}°+\mathrm{y}}{\mathrm{CH}}\Rightarrow\mathrm{CH}=\frac{\sqrt{\mathrm{2}}\mathrm{xsin}\left(\mathrm{2y}\right)}{\left(\mathrm{cosy}−\mathrm{siny}\right)} \\ $$$$\frac{\mathrm{CH}}{\mathrm{CD}}=\sqrt{\mathrm{2}}\mathrm{cosy};\:\frac{\mathrm{HE}}{\mathrm{ED}}=\sqrt{\mathrm{2}}\mathrm{cosy} \\ $$$$\frac{\mathrm{CH}}{\mathrm{CD}}=\frac{\mathrm{HE}}{\mathrm{ED}}\:\Rightarrow\:\mathrm{HD}\:\mathrm{bisects}\:\angle\mathrm{CHE}\:\Rightarrow\:\angle\mathrm{CHD}=\frac{\mathrm{90}°}{\mathrm{2}} \\ $$$$\Rightarrow\angle\mathrm{CHD}=\mathrm{45}° \\ $$
Answered by fantastic2 last updated on 15/Nov/25
Commented by fantastic2 last updated on 15/Nov/25
big mistake.
$${big}\:{mistake}. \\ $$
Answered by mr W last updated on 16/Nov/25
Commented by Mingma last updated on 16/Nov/25
Perfect ��
Commented by mr W last updated on 16/Nov/25
say AB=1 BD=(√2) ((BH)/(cos 2α))=(1/(cos (45°−α)))=BE ⇒BH=((cos 2α)/(cos (45°−α)))=((cos^2 α−sin^2 α)/(cos (45°−α))) =(((cos α−sin α)(cos α+sin α))/(((1/( (√2))) cos α+(1/( (√2))) sin α))) =(√2) (cos α−sin α) tan θ=((DK)/(HK))=((DK)/(BK−BH)) =(((√2) sin α)/( (√2) cos α−(√2) (cos α−sin α))) =(((√2) sin α)/( (√2) sin α))=1 ⇒θ=45°
$${say}\:{AB}=\mathrm{1} \\ $$$${BD}=\sqrt{\mathrm{2}} \\ $$$$\frac{{BH}}{\mathrm{cos}\:\mathrm{2}\alpha}=\frac{\mathrm{1}}{\mathrm{cos}\:\left(\mathrm{45}°−\alpha\right)}={BE} \\ $$$$\Rightarrow{BH}=\frac{\mathrm{cos}\:\mathrm{2}\alpha}{\mathrm{cos}\:\left(\mathrm{45}°−\alpha\right)}=\frac{\mathrm{cos}^{\mathrm{2}} \:\alpha−\mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{cos}\:\left(\mathrm{45}°−\alpha\right)} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{cos}\:\alpha−\mathrm{sin}\:\alpha\right)\left(\mathrm{cos}\:\alpha+\mathrm{sin}\:\alpha\right)}{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{cos}\:\alpha+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{sin}\:\alpha\right)} \\ $$$$\:\:\:\:\:\:\:\:=\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\alpha−\mathrm{sin}\:\alpha\right) \\ $$$$\mathrm{tan}\:\theta=\frac{{DK}}{{HK}}=\frac{{DK}}{{BK}−{BH}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\alpha}{\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:\alpha−\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\alpha−\mathrm{sin}\:\alpha\right)} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\alpha}{\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:\alpha}=\mathrm{1} \\ $$$$\Rightarrow\theta=\mathrm{45}° \\ $$

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