Question

"Question 1 ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.

Class 9 - Math - Triangles Page 133"

Answer 1

[fig. is in the attachment]


Let ABC be a triangle draw perpendicular bisectors of BC and AC which intersect each other at O. Thus, OM & ON are the perpendicular bisectors of sides BC & AC of ∆ABC.

So, O is equidistant from two end points B & C of line segment BC, as O lies on the perpendicular bisector of BC. Similarly, O is equidistant from C & A.

Hence, O is an orthocentre of ∆ABC.

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Hope this will help you........

Answer 2

Hi friend..

Let OD and OE be the perpendicular bisectors of sides BC and CA of ∆ABC.

So that , O is a equidistant from two ends B and C of line segment BC as O lies on the perpendicular bisectors of BC.
Same,O is equidistant from C and A.

Thus, the point of interaction O of the perpendicular bisectors of sides BC,CA,and AB is the required point..

Hope it is helpful