prove that : $MN=PQ=AC$

Question

Found this problem in book with answer but no detail! $ABCD$ be a convex quadrilateral $ABM,CDP,BCN,ADQ$ equilateral triangle such that $M,P$ inside $ABCD$ and $N,Q$ outside $ABCD$ then prove that :

$1)$ $MN=PQ=AC$

$2)$ what we can say about quadrilateral $MNPQ$

Answer :

$1)$ use rotational

$R(B,\frac{π}{3}),R(D,\frac{π}{3})$

2) if $MN//PQ$ (Parallel) then $MNPQ$ parallelogram or the points $M,N,P,Q$ are Colinear .

Now i don't understand rotational can we solve without rotational ? And what about second question ?

Answer 1

1) Rotate the triangle PDQ by 60 degrees around D. It is obvious that the triangles PDQ and CDA are equivalent. In the same way, show that ABC and MBN are equivalent.

2) From 1), MN=PQ. If MN and PQ are parallel, than the triangles PMQ and MPN are equivalent. Therefore either MNPQ is a parallelogram or M, N, P, Q are colinear.

Answer 2

More hint.

Rotation of a vector in a plane by $\alpha$ around a point of the plane it's a rotation of the vector by $\alpha$ around a tail of the vector.

Answer 3

If you don't like rotating triangle $MBN$ about $B$ to become triangle $ABC$, then you can just prove that those two triangles are congruent: you have $MB=AB,BN=BC$ and $\angle MBN=\angle MBC+60^\circ=\angle ABC$.

Similarly for the rotation about $D$ to establish that $PQ=AC$.

For the second part, $MNPQ$ is a parallelogram. You have already established that $MN=PQ$ in the first part. They are parallel, because rotating $AC$ through $60^\circ$ one way gives $MN$ and through $60^\circ$ the other way gives $PQ$.

You should get used to using rotations, they can be a powerful technique.