Question

In the given figure, $\Delta AEC$ and $\Delta DBF$ are equilateral. Prove that all other triangles are equilateral. (Given that bases of the triangles are parallel).
[4 MARKS]

Answer 1

Steps: 3 Marks
Proof: 1 Mark

As given in question $\Delta AEC$ and $\Delta DBF$ are equilateral so angle made at vertexes $={60}^{\circ }$

$\angle AEC=\angle AMB={60}^{\circ }$ (Corresponding Angles) because $FB\parallel EC$.

Similarly $\angle ACE=\angle ANM={60}^{\circ }$ (Corresponding Angles).

For $\Delta AMN$, from above two conditions, we find out the remaining two angles $\angle AMB$ and $\angle ANM={60}^{\circ }$

Hence all angles of $\Delta AMN$ are equal to ${60}^{\circ }$.

Now consider $\Delta BNO,\angle FBD={60}^{\circ }$ (it is the vertex of equilateral triangle) and $\angle BNO={60}^{\circ }$ (vertically Opposite Angles)

$\angle BON={60}^{\circ }$ (Angle sum property of triangle). Hence $\Delta BNO$ is also equilateral.

Similarly, we can prove it for the remaining triangle.