SUBIECTUL IV

In the plane of triangle $A B C$, with $A B=A C=a, B C=a \sqrt{2}$, a perpendicular $D C$ is raised such that $D C=a$

a) Calculate $m[\square((D A C),(D B C))]$; b) If $M$ and $E$ are the midpoints of segments $[B C]$ and $[D A]$ respectively, prove that point $M$ is equidistant from points $A, B, C$ and $E$; c) Determine $\operatorname{tg}(\square(A C, B D))$.

## BAREM CLASA A VIII-A
## SUBJECT IV

a) $A C, B C \perp D C \Rightarrow m[\square((D A C),(D B C))]=m(\square B C A)(1 \text{p})$;

$\square A B C$ is isosceles and right-angled at $A \quad m(\square B C A)=45^{\circ}(1 \text{p})$;
b) $[A M]$ is the median in $\square A B C$, right-angled at $A \Rightarrow M A=M B=M C=\frac{a \sqrt{2}}{2}(\mathbf{0}, \mathbf{5 p})$; $A M \perp B C, A M \perp D C \Rightarrow A M \perp(D B C)$ (1p); [ME] is the median in $\square D M A$, right-angled at $M D M=\frac{A D}{2}=\frac{a \sqrt{2}}{2}(\mathbf{0}, \mathbf{5 p})$;
c) Let $P$ be the midpoint of $[A B]$. It is demonstrated that: $P M \square A C, P E \square B D \Rightarrow m(\square A C, B D)=m(\square E P M)(1 \text{p}) ;$ $B D=a \sqrt{3}, P E=\frac{a \sqrt{3}}{2}, P M=\frac{a}{2}$ and, since $M E=\frac{a \sqrt{2}}{2} \Rightarrow \square M E P$ is right-angled at $M$ (1p); $\operatorname{tg} \square E P M=\sqrt{2}$ (1p).

![](https://cdn.mathpix.com/cropped/2024_06_07_0736c21030929a24a3b4g-3.jpg?height=688&width=557&top_left_y=684&top_left_x=378)

$\mathrm{N}$

$\mathrm{C}$

B

![](https://cdn.mathpix.com/cropped/2024_06_07_0736c21030929a24a3b4g-3.jpg?height=639&width=625&top_left_y=797&top_left_x=1087)

B