Cho tam giác ABC vuông tại A. Gọi I là tâm đường tròn nội tiếp tam giác ABC.

Đáp án A

Gọi r là bán kính đường tròn nội tiếp tam giác ABC.

Ta có:$IA=\frac{r}{\sin \frac{A}{2}};IB=\frac{r}{\sin \frac{B}{2}};IC=\frac{r}{\sin \frac{C}{2}}$.

Vế trái $\frac{1}{x^2}=\frac{{\sin }^2\frac{A}{2}}{r^2}=\frac{{\sin }^{2}45°}{r^2}=\frac{1}{2r^2}$.

Vế phải:

 $\begin{array}{l}\frac{1}{y^2}+\frac{1}{z^2}+\frac{\sqrt{a}}{yz}=\frac{{\sin }^2\frac{B}{2}}{r^2}+\frac{{\sin }^2\frac{C}{2}}{r^2}+\frac{\sqrt{a}.\sin \frac{B}{2}.\sin \frac{C}{2}}{r^2}\\ =\frac{1}{r^2}\left[\frac{1-\cos B+1-\cos C}{2}+\frac{\sqrt{a}.\left(\cos \frac{B-C}{2}-\cos \frac{B+C}{2}\right)}{2}\right]\\ =\frac{1}{2r^2}\left[2-2\cos \frac{B+C}{2}\cos \frac{B-C}{2}+\sqrt{a}\cos \frac{B-C}{2}-\sqrt{a}\cos \frac{B+C}{2}\right]\\ =\frac{1}{2r^2}\left[2-2\cos 45°.\cos \frac{B-C}{2}+\sqrt{a}\cos \frac{B-C}{2}-\sqrt{a}\cos 45°\right]\\ =\frac{1}{2r^2}\left[2-2.\frac{\sqrt{2}}{2}.\cos \frac{B-C}{2}+\sqrt{a}\cos \frac{B-C}{2}-\sqrt{a}.\frac{\sqrt{2}}{2}\right]\\ =\frac{1}{2r^2}\left[2-\frac{\sqrt{2a}}{2}-\left(\sqrt{2}-\sqrt{a}\right)\cos \frac{B-C}{2}\right]\end{array}$

Do $\frac{1}{x^2}=\frac{1}{y^2}+\frac{1}{z^2}+\frac{\sqrt{a}}{yz}\Rightarrow \frac{1}{2r^2}=\frac{1}{2r^2}\left[2-\frac{\sqrt{2a}}{2}+\left(\sqrt{a}-\sqrt{2}\right)\cos \frac{B-C}{2}\right]\Rightarrow \left\{\begin{array}{l}2-\frac{\sqrt{2a}}{2}=1\\ \sqrt{a}-\sqrt{2}=0\end{array}\right.\Rightarrow a=2$.