metafor wrote: Consider a triangle $ ABC$, with $ \angle A = 90^{\circ}$, and $ AC > AB$. Let $ D$ be a point in $ AC$ such that $ \angle ACB = \angle ABD$. Draw an altitude $ DE$ in triangle $ BCD$. If $ AC = BD + DE$, find $ \angle ABC$ and $ \angle ACB$. $ \angle ACB=\angle ABD=\beta$ > $ \angle DBE=90^{o}-2\beta$ > $ \bigtriangleup BDE$ ----> $ sin(90^{o}-2\beta)=\frac{DE}{BD}=cos2\beta$ ----> $ DE=BDcos2\beta$ $ \bigtriangleup ABD$ ----> $ cos\beta=\frac{AB}{BD}$- > $ BD=\frac{AB}{cos\beta}$- > $ DE=\frac{ABcos2\beta}{cos\beta}$ $ AC=BD+DE=BD=\frac{AB}{cos\beta}+\frac{ABcos2\beta}{cos\beta}=\frac{AB(1+cos2\beta)}{cos\beta}=\frac{AB(1+2cos^{2}\beta-1)}{cos\beta}=2ABcos\beta$ ----> $ \frac{AC}{AB}=2cos\beta$ $ \bigtriangleup ABC$ > $ cot\beta=\frac{AC}{AB}$ > $ \frac{AC}{AB}=2cos\beta=cot\beta$ > $ sin\beta=\frac{1}{2}$ ----> $ \beta=30^{o}$

metafor wrote: Consider $ \triangle ABC$ with $ A = 90^{\circ}$ and $ b > c$ . Denote $ \left\|\begin{array}{c} D\in (AC)\ ,\ m(\widehat {ABD})=C\\\\ E\in (BC)\ ,\ DE\perp BC\end{array}\right\|$ . Prove that $ AC = BD + DE\ \Longleftrightarrow\ a=2c$ . Proof. Denote $ F\in (BC)$ for which $ AF\perp BC$ . Observe that $ b^2+c^2=a^2$ and $ AF=\frac {bc}{a}$ . From $ \triangle ABD\ \sim\ \triangle ACB$ obtain $ \frac cb=\frac {BD}{a}=\frac {DA}{c}$ , i.e. $ BD=\frac {ac}{b}$ and $ DA=\frac {c^2}{b}$ . Thus, $ DC=AC-AD$ , i.e. $ DC=\frac {b^2-c^2}{b}$ . Since $ DE\parallel AF$ obtain $ \frac {CD}{CA}=\frac {DE}{AF}$ $ \implies$ $ \frac {b^2-c^2}{b^2}=\frac {DE}{\frac {bc}{a}}$ , i.e. $ DE=\frac {c(b^2-c^2)}{ab}$ . Therefore, $ \boxed {\ AC=BD+DE\ }$ $ \Longleftrightarrow$ $ b=\frac {ac}{b}+\frac {c(b^2-c^2)}{ab}$ $ \Longleftrightarrow$ $b^2(a-c)=c(a^2-c^2)$ $ \Longleftrightarrow$ $ \boxed{b^2=c(a+c)}\ (*)$ $ \Longleftrightarrow$ $ a^2-c^2=c(a+c)$ $ \Longleftrightarrow$ $ a-c=c$ $ \Longleftrightarrow$ $a=2c\iff$ $B=60^{\circ}$ . Remark. From the relation $(*)$ , i.e. $b^2=c(a+c)$ obtain shorterly that $B=2C$ , i.e. $B=60^{\circ}$ .

You're enjoying this, aren't you?

So-called ... middling. And yet I like it.

$AB$ is tangent to the circle $\odot (BCD)$, hence $AB^2=AD\cdot AC \ (\ 1\ ), BADE$ is cyclic, consequently $\widehat{AED}=\widehat{ABD}=\widehat{BCA}$ and $AE$ is tangent to the circle $\odot (CDE)\implies AE^2=AD\cdot AC\ (\ 2\ )$. From $(1)$ and $(2)\implies AE=AB \iff\angle CDE=\angle ADB\ (\ 3\ )$. Extend $(BD$ with $DF=DE$; from the given relation we get $BF=AC$ and, with $(3)$ we infer $\triangle DFC\equiv\triangle DEC$, hence $CF=CE$ and $DFCE$ is cyclic and $\widehat{DFE}=\widehat{DCE}$. We conclude that $\triangle BEF\equiv\triangle AEC$ (a.s.a), hence $EF=CE$, or triangle $CEF$ is equilateral and the result follows. Best regards, sunken rock

well i don't think i have a good solution like that of zaya_yc...!! But yea this is one..!! $since \ \angle CDB = 90^\circ + \theta \ and \ \angle CDE = 90^\circ - \theta \ so \ \angle EDB = 2\theta$ we draw DM || AB so $\angle EDM = \angle BDM = \theta$ By Thales' theorem, $\frac {DM}{AB} = \frac {CD}{AC}$ $so, \ \frac {q sec\theta}{p cos\theta} = 1 - \frac {p sin\theta}{(p+q)}$ $or, \ \frac {q}{p cos^2\theta} + \frac {p sin\theta}{(p+q)} = 1$ $\implies pq + q^2 + p^2cos^2\theta.sin\theta = p^2cos^2\theta + pq cos^2\theta$ $\implies pqsin^2\theta + p^2cos^2\theta(sin\theta - 1) + q^2 = 0 ----- (i)$ $Now, \ clearly \ \triangle CDE \sim \triangle BDA$ $so, \ \frac {BD}{AD} = \frac {CD}{DE}$ $=> \frac {p}{p sin\theta} = \frac {p + q - psin\theta}{q}$ $\implies pq = p^2 sin\theta + pq sin\theta - p^2 sin^2\theta$ $\implies pq (1 - sin\theta) = p^2 sin\theta(1 - sin\theta)$ $=> q = psin\theta ------ (ii)$ putting q from (ii) in (i) we get, $p^2 sin^3 \theta + p^2 cos^2\theta (sin\theta - 1) + p^2 sin^2 \theta = 0$ $\implies sin^3\theta + sin\theta - sin^3 \theta - cos^2\theta + sin^2 \theta = 0$ $\implies 2 sin^2\theta + sin\theta - 1 = 0$ $\implies 2 sin^2\theta + 2sin\theta - sin\theta - 1 = 0$ $\implies (sin\theta + 1)(2 sin\theta - 1) = 0$ $rejecting\ sin\theta = -1 \ we \ get, \ sin\theta = 1/2 \ or, \ \theta = \pi/6$