|AD-CD|=side of ABC, <A=C/2 (2017 All-Siberian Correspondence 8.2)

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|AD-CD|=side of ABC, <A=C/2 (2017 All-Siberian Correspondence 8.2)

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#1 h May 9, 2021, 11:42 AM

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In triangle $ABC$ , angle $A$ is half the angle $C$ , and point $D$ on side $AC$ is the foot of the altitude from $B$ . Prove that the difference of the line segments into which $D$ divides $AC$ , equals to one of the sides of triangle $ABC$ .

This post has been edited 1 time. Last edited by parmenides51, May 9, 2021, 7:45 PM
Reason: title typo

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#2 h May 9, 2021, 7:38 PM

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Rule of sines in $\triangle BCD\ :\ \frac{CD}{\cos \gamma}=\frac{BD}{\sin \gamma}=BC$ .
Rule of sines in $\triangle ABD\ :\ \frac{AD}{\cos \frac{\gamma}{2}}=\frac{BD}{\sin \frac{\gamma}{2}}=AB$ .
Rule of sines in $\triangle ABC\ :\ \frac{AB}{\sin \gamma}=\frac{BC}{\sin \frac{\gamma}{2}}$ .

We eliminate $BD\ :\ \frac{CD}{\cos \gamma}=\frac{AD \sin \frac{\gamma}{2}}{\sin \gamma\cos \frac{\gamma}{2}}$ .
$\frac{CD}{AD}=\frac{\cos \gamma \sin \frac{\gamma}{2}}{\sin \gamma\cos \frac{\gamma}{2}}$ ,
$\frac{AD-CD}{AD}=\frac{\sin \gamma\cos \frac{\gamma}{2}-\cos \gamma \sin \frac{\gamma}{2}}{\sin \gamma\cos \frac{\gamma}{2}}=\frac{\sin \frac{\gamma}{2}}{\sin \gamma\cos \frac{\gamma}{2}}$ ,

$AD-CD=\frac{\sin \frac{\gamma}{2} \cdot AD}{\sin \gamma\cos \frac{\gamma}{2}}=\frac{\sin \frac{\gamma}{2} \cdot AB \cos \frac{\gamma}{2}}{\sin \gamma\cos \frac{\gamma}{2}}$ ,
$AD-CD=\frac{AB}{2\cos \frac{\gamma}{2}}=\frac{BC \sin \gamma}{2\cos \frac{\gamma}{2}\sin \frac{\gamma}{2}}=BC$

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#3 h May 10, 2021, 5:53 AM

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Taking $E$ reflection of $C$ in $D$ you get $AE=BE=BC$ , but $AE=AD-CD$ , done.

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