03/10/2023

In $\triangle{ABC}$, $\angle{BAC}=45^{\circ}, AD \perp BC, BD=3, AD=9$, find $CD$ .

Solve 1:

\(\begin{multline} \nonumber \shoveleft \text{ Make } BE\perp AC, \text{ connect } DE, \text{ easy to see }\\ \shoveleft AB=3\sqrt{10}, AE=BE=\dfrac{AB}{\sqrt{2}}=3\sqrt{5}\\ \shoveleft \text{By the Ptolemy's theorem we know: }\\ \shoveleft AB \cdot DE+ BD \cdot AE = BE \cdot AD\\ \shoveleft \implies DE=3\sqrt{2}, \text{ let } CD=x, CE=y\\ \shoveleft \triangle{CDE}\sim\triangle{CAB} \implies \dfrac{DE}{AB}=\dfrac{CD}{CA}=\dfrac{CE}{CB}\\ \shoveleft \implies \dfrac{1}{\sqrt{5}}=\dfrac{x}{3\sqrt{5}+y}=\dfrac{y}{3+x}\\ \shoveleft \implies y=\dfrac{3\sqrt{5}}{2}, x=\bbox[5px, border: 1px solid black]{\dfrac{9}{2}} \end{multline}\) Solve 2: \(\begin{multline}\nonumber \shoveleft \text{ Make } BE\perp AC, \text{ connect } DE, \text{ easy to see }\\ \shoveleft AB=3\sqrt{10}, AE=BE=\dfrac{AB}{\sqrt{2}}=3\sqrt{5}\\ \shoveleft \implies [ABC]=\dfrac{AD \cdot BC}{2}=\dfrac{9(3+CD)}{2}\\ \shoveleft =\dfrac{BE \cdot AC}{2}=\dfrac{3\sqrt{5}(\sqrt{81+CD^2})}{2}\\ \shoveleft \implies CD=\bbox[5px, border: 1px solid black]{\dfrac{9}{2}} \end{multline}\)