Semicircle with tangent and perpendicular: prove that DE·BC = CD·R

If you draw the whole circle, the relationship becomes clearer.

Circle $\odot O$ is inscribed in isosceles $\triangle ECE'$. Let $E'C = EC = a$, and $EE' = c$. Furthermore, let $BC = x$. Then $|\triangle ECE'| = rs$, where $s = (2a+c)/2 = a + c/2$ is the semiperimeter and $r$ is the inradius. But we also have $|\triangle ECE'| = EA \cdot AC = (c/2)(2r + x)$. So $$(2r + x)c = 2rs$$ or $$\frac{c}{2} \cdot x = (s-c)r.$$ But since $s = CD + DE + EA$ and $DE = EA = c/2$, we have $CD = s - c$. Therefore, $$DE \cdot BC = CD \cdot r,$$ as required.

Let $\angle EOA$ and $\angle EOD$ be $\theta$.

$\angle DOB+\angle OBD+\angle ODB=\pi$

$\therefore\angle DOB+2\angle OBD=\pi$

$\therefore\angle OBD=\frac{\pi-\angle DOB}{2}=\frac{\pi-(\pi-2\theta)}{2}=\theta$

$\therefore EO \parallel DB$

$\therefore$ In $\triangle EOC$, $\frac{CD}{DE}=\frac{CB}{BO}$

$\therefore DE\cdot BC=CD\cdot r$

Let $2\vartheta=\angle BOD$ $$ \begin{align} \tag1\tan2\vartheta&=\frac{CD}{OD}=\frac{CD}R\\ \tag2CD&=R\tan2\vartheta\\ \tag3OC&=R\sec2\vartheta\\ \tag4BC&=R\left(\sec2\vartheta-1\right)\\ \tag5\angle AOD&=\pi-2\vartheta\\ \tag6\angle EOD&=\frac12\angle AOD=\frac\pi2-\vartheta\\ \tag7DE&=R\tan\angle EOD\\ \tag8\frac{DE\cdot BC}{R^2}&=\tan\angle EOD\left(\sec2\vartheta-1\right)\\ \tag9&=\tan\left(\frac\pi2-\vartheta\right)\left(\frac1{\cos2\vartheta}-1\right)\\ \tag{10}&=\cot\vartheta\left(\frac{1-\cos2\vartheta}{\cos2\vartheta}\right)\\ \tag{11}&=\frac{\cos\vartheta}{\sin\vartheta}\left(\frac{2\sin^2\vartheta}{\cos2\vartheta}\right)\\ \tag{12}&=\frac{2\sin\vartheta\cos\vartheta}{\cos2\vartheta}\\ \tag{13}&=\frac{\sin2\vartheta}{\cos2\vartheta}\\ \tag{14}&=\tan2\vartheta\\ \tag{15}&=\frac{CD}R\qquad\text{(proved)} \end {align} $$

$\triangle AEC\sim \triangle OCD$

$\frac {EC}{OC}=\frac{ED}{OD}\Rightarrow \frac {EC}{ED}=\frac{OC}{OD}$

$\frac{EC-ED}{ED}=\frac{OC-OD}{OD}$

$OD=R$

$OC-OD=BC$

$EC-ED=DC$

$ED=AB$

$\frac{DC}{ED}=\frac {BC}R$

$DE\cdot BC= CD \cdot R$

My Proof

Draw the segments BD, AD, and OE.

Step 1: The line OE is the radical line from point E to the circle (O, R). By a well-known theorem, this line is the perpendicular bisector of chord AD. Therefore: $$OE \perp AD \quad (1)$$

Step 2: Since angle ∠ADB is inscribed in a semicircle, we have: $$\angle ADB = 90°$$

Therefore: $$BD \perp AD \quad (2)$$

Step 3: From (1) and (2), since both OE and BD are perpendicular to AD: $$BD \parallel OE$$

Step 4: By Thales' theorem, points D and B divide segments CE and CO proportionally: $$\frac{CD}{DE} = \frac{BC}{OB} = \frac{BC}{R}$$

Cross-multiplying: $$DE \cdot BC = CD \cdot R \quad \square$$

Note: If ∠ACD = 30°, then the conclusion becomes: $$CD \cdot BC = DE \cdot R = \sqrt{3} \cdot R^2$$