#### Question

In the figure, \triangle 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.

Find the length of CM when \triangle MPN is a right triangle

If \angle MPN rotates around point P, find the length of CN\cdot AM

Collected in the board: Triangle

Steven Zheng posted 5 days ago

Drag the slide to observe how the triangle MPN changes

Case 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 \triangle PAM to the sum of the two interior opposite angles.

\angle PMA = \angle PCM + \angle CPM

Therfore, \angle PCM = \angle CPN

Now for \triangle CPN and \triangle AMP , there are 2 pairs of equal angles

\triangle CPN \sim \triangle 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

Steven Zheng posted 5 days ago

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