[swastik pramanik]

Three equal circles pass through a given point H and meet one another two by two at A, B, C. Prove that H is the orthocentre of \(\Delta ABC\)?

(Source: From Challenge and Thrills of Pre-College Mathematics; Pg No.-97; Q. No. - 16 of EXERCISE 4.2)

[Dr. Ashani Dasgupta]

Swatik,

Here are some hints. First look at the diagram:

Note that orthocenter is the intersection point of the altitudes of a triangle.

We wish to show \( AH \perp BC \) (and similarly \(BH \perp AC \) etc. ).

Now can you show AH is perpendicular to a segment parallel to BC? It is clearly marked in the picture but you should think closely why that segment is parallel to BC?

One way to do it is to show that segment and BC are part of a parallelogram. (Note that opposite sides of a parallelogram are equal and parallel).

Finally, can you find a circle of which H is a center? That circle is not drawn in the picture above.

Bonus: This is just for fun. In the picture, we have a hexagon that is equilateral but not necessarily equiangular. This is quite different from what happens with triangles (where equilateral means equiangular).

What can you, in general, say about angles of an equilateral polygon? This is not required for solving this problem. But you may think about it.

[swastik pramanik]

sir i am not able to prove that O1O2 || BC. please give some hint.

sir i think angles can be anything in equilateral polygon.

[Subrata Ghosh]

Hello Swastik,

The image below can give you some hint you are looking for.

[swastik pramanik]

Join \(AO_3\). now \(BO_3||O_1H\)...and \(O_1H||AO_2\).....so \(BO_3||AO_2\)....simmilaly, \(AO_1||CO_3\)...so, \(\angle O_1AO_3=\angle AO_3C=x\)...and \(\angle O_2AO_3=\angle AO_3B=y\)....so \(\angle O_1AO_2=\angle BO_3C=x+y\).....then \(\Delta AO_1O_2 \cong \Delta BO_3C\).....and we are done.....Am I correct SIr????

[Author]

Posts