Help me with the diagram

[swastik pramanik]

In triangle ABC, AB=AC and A is also nine point centre of triangle ABC.

Can someone help me with the diagram?

[Sankhadip Chakraborty]

Hi,

Note that the nine-point centre lies halfway between the circumcentre and the orthocentre. If both of them lie inside the triangle, this cannot happen. Hence, at least one of them has to lie outside. This means that the triangle has to be obtuse (a right-angled triangle does not work because in that case, the orthocentre coincides with a vertex).  Hence, the image you should have in mind is that of an obtuse triangle with the obtuse angle being A.

[Nitin Prasad]

 

Let D, E, F be midpoints of sides BC, CA, AB respectively. Hence the circle in the above figure is the nine-point circle of triangle ABC

If vertex A is the nine-point center of triangle ABC then following are some of the observations

• AE=AF=r (radius of the circle in the above figure). Therefore AB=AC=2r, hence this condition was redundant.
• AD=r. Since triangle ABC is isosceles, therefore AD is also an altitude, i.e. $$\angle ADC=90^\circ$$
• $$\frac{AD}{AB}=\frac{1}{2}\Longrightarrow \angle BAD=60^\circ$$.

Also observe that whenever in a $$\triangle XYZ, \angle  YXZ=120^\circ \& XY=XZ$$, then X is a nine-point center of triangle XYZ

Hence A is a nine-point center of a triangle ABC if and only if-$$\angle BAC=120^\circ \& AB=AC$$

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