Question

In $\Delta ABC,\angle A={45}^{\circ },\angle B={60}^{\circ }$, and $\angle C={75}^{\circ }$. Find the angles formed by joining the mid-points of the sides of the given triangle.

Answer 1

Consider the figure below:

$D,E$ and $F$ are the mid-points of the sides $BC,AC$ and $AB$ respectively.

From the converse of mid-point theorem, we can say that

$\Rightarrow DE\parallel AB,DF\parallel AC$ and $EF\parallel AB$

Also, $DE=\frac{AB}{2}=AF,DF=\frac{AC}{2}=AE$ and $EF=\frac{AB}{2}=BD$

Thus, $AFDE,FBDE$ and $EFDC$ are parallelograms.

In $△ABC$ and $△DEF$,
$\angle EDF=\angle CAB={45}^{\circ }$ [Opposite angles of parallelogram, $AFDE$]
$\angle DEF=\angle ABC={60}^{\circ }$ [Opposite angles of parallelogram, $FBDE$]
and $\angle EFD=\angle ACB={75}^{\circ }$ [Opposite angles of parallelogram, $EFDC$]
Hence, the angles of the triangle formed by joining the mid-points of the sides of $△ABC$ are ${45}^{\circ },{60}^{\circ },and{75}^{\circ }.$