In a right-angled triangle ABC (∠C=90°), the height CD equal to h is drawn.
The points M and N are the midpoints of the segments BD and AD.
Find the distance from the point C to the orthocenter of the triangle CMN (Fig. 2).
In right-angled triangle ABC with right angle at C, height CD (which is also the altitude from C to AB) divides AB into AD and DB. Since M and N are the midpoints of BD and AD respectively, triangle CMN will have angles that are congruent to those of triangle CAB (since they share two angles and all the third angles are right angles). This makes triangle CMN similar to triangle CAB.
In a right-angled triangle, the orthocenter is the vertex where the right angle is located. So, for triangle CMN, the orthocenter is point C itself. Therefore, the distance from point C to the orthocenter of triangle CMN is 0.
a geometric view:
A geometric transformation which combines
1) rotation around point D at 90 degrees clockwise
followed by
2) homothety with center D and ratio |DB|/|DC|,
such transformation maps triangle ADC to triangle CDB.
In particular, the median AK maps to the median CM (where AK goes from A to DC; K is the midpoint of DC). Thus, the angle between AK and CM is the angle of rotation 90 degrees.
Hence, AK is parallel to NQ, and NQ is a midsegment in triangle ADK.
Combining (i) Q is midpoint of DK, and (ii) K is midpoint of DC, gives the length (3/4)h for CQ.
Steftcho P. Dokov A very interesting solution to the problem!
I am delighted!
Thanks. Interesting, and challenging geometry-questions,
you are posting as well, Liudmyla Hetmanenko
AS orthogonal on CM with S on CM; R the intersection point between AS and CD.
NQ//AR and AN=DN=> RQ=DQ
angle CBM=angle ACR and angle CAR=angle BAM( = 90 - MCA) => triangles BCM and CRA are similar => CR/BM=AC/BC => CR=(BD)(AC)/(2BC)
BC2 =(AB)(BD) implies now CR=(AC)(BC2)/(AB)(2BC)=(AC)(BC)/(2AB)=h/2
Now it results CQ=h/2+h/4