Similar triangle proof in parallelogram

Question

Can anyone help me with this task.
From the top of a parallelogram $ABCD$ lowered the vertical $AM$ and $AN$ on the lines BC and CD . Prove that triangles $\triangle ABC$ and $\triangle AMN$ similar .

Answer 1

The first thing is to show that A, M N, C are con-cyclic.

Added:-

If $\angle B = \theta$, then $\angle BAM = 90^0 - \theta$ and therefore $\angle MAN = \theta$ because of complementary angles.

Note that $\angle BCN = \theta$ also because of alternate angles.

“$\angle MAN = BCN$” implies $MACN$ is a cyclic quadrilateral [converse of angles in the same segement].

Therefore, $\angle ACM = \angle ANM$ and the reason is they are angles in the same segment.

By AA, the two said triangles are thus similar.

Answer 2

It is easily seen that the red and the blue $\beta$s in the following figure are equal.

On the other hand, since $AM$ is the height of the parallelogram $P:=ABCD$ with respect to the base $BC$, and $AN$ is the height of $P$ with respect to the base $CD$, or $BA$, we have $$|BC|\cdot|AM|={\rm area}(P)=|BA|\cdot|AN|\ ,$$ and therefore $${|AN|\over|AM|}={|BC|\over|BA|}\ .$$ Taken together the claim follows.

Answer 3

Consider the $\square AMNC $ $$\angle AMC = \angle ANC$$ This means, $\square AMNC$ is a cyclic quadrilateral. Also, as $\angle AMC=90^0$, AC is the diameter of the circle.

As a property of a cyclic quadrilateral, $\angle ANM = \angle ACM = \angle ACB $ -- Statement 1 (internal angle property)

Also, $\angle ABC = \angle BCN$ (alternate angles)

and $\angle MAN = \angle MCN$ (internal angles of cyclic quadrilateral)

Hence, $\angle ABC = \angle MAN$ -- Statement 2 From Statement 1 and Statement 2, we get, $\triangle ABC$ is similar to $\triangle MAN$. Hence, proved.