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?

Answer 1

ANSWER
...............................Angle BNC = CMB   
$=$

 
$=$
=  angles in the same segment of the circle  are equal .


= BC is a chord and 

THE


 B, C, M, N all lie on a circle 


 = therefore they  are concyclic

⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃⚃
A, B and C are three non-collinear points on the circle. Let O be the centre of the circle.

Now, the sides AB, BC and AC of ∆ABC are the chords of the circle.

 

It is known that the perpendicular bisector of a chord of a circle passes through the centre of that circle.

Thus, the perpendicular bisector of each chords AB, BC and AC passes through O.

This shows that the perpendicular bisectors of AB, BC and AC intersect at O.

Thus, the perpendicular bisectors of AB, BC and AC are concurrent .


$= = = = = = = = = = =$

Answer 2

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.