In the figure, $$ riangle ABC$$ is a right triangle such that $$ \angle C=90\degree $$， $$ BC=6$$， $$ AC=8$$. Point P is the midpoint of $$AB$$. Taking point P as the vertex, construct $$\angle MPN$$ so that $$ \angle MPN=\angle A$$ and the arms of $$\angle MPN$$ intersect AC at points M, N respectively.

Question

In the figure, $$	riangle ABC$$ is a right triangle such that  $$ \angle C=90\degree $$，  $$ BC=6$$，  $$ AC=8$$. Point P is the midpoint of $$AB$$. Taking point P as the vertex, construct   $$\angle MPN$$ so that $$  \angle MPN=\angle A$$ and the arms of $$\angle MPN$$ intersect AC at points M, N respectively.

Uniteasy Admin · Accepted Answer

Drag the slide to observe how the triangle MPN changesCase 1, if $$\angle PMN = 90\degree $$,
then $$PM\parallel BC $$. Since point P is the midpoint of AB，point M is also the midpoint of AC. Therefore, the length of CM is the half of AC.
$$CM= 4$$ Connect point P to point C. 
Since P is the midpoint of hypotenuse of a right triangle, 
$$PC=PA$$
Then,
$$\angle PCM = \angle A = \angle MPN$$
The external angle of $$	riangle PAM$$  to the sum of the two interior opposite angles.
$$\angle PMA = \angle PCM + \angle CPM$$
Therfore, $$\angle PCM  = \angle CPN$$
Now for $$	riangle CPN$$ and  $$	riangle AMP$$ , there are 2 pairs of equal angles
 $$	riangle CPN \sim 	riangle AMP$$
For similar triangles, the ratios of corresponding sides are equal 
$$ \dfrac{CN}{CP} = \dfrac{PA}{AM}  $$
Cross multiply
$$CN\cdot AM = PA\cdot CP=25$$
Therefore, the product of CN and AM is a constant

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