Question

in trianglr ABC BM is perpendicular to AC and CN ​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: Given :-$

Total question :

If BM and CN are the perpendiculars drawn on the sides AB and AC of a triangle ABC, then prove that the points B,C, M and N are cyclic?

$\: \: \: \: \: \: \: \: \: Solution :-$

In the above figure, BM & CN are perpendiculars to AC & AB respectively.

As < BNC = 90° (given that CN perpendicular to AB of tri ABC)

So, we can say that points B, N, C lie on a semi circle, the diameter of which is BC. ( as angle on a semicircle= 90°)

Now, < BMC= 90° ( given that BM perpendicular to AC of tri ABC

So, points B, M, C too lie on a semicircle the diameter of which is the same BC.

So, diameter is the same for both the semicircles.

That shows that B,C, M, N are lying on the same semi circular arc. So these points are concyclic.