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प्रश्न
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
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उत्तर
The point which is equidistant from all the sides of a triangle is called the incentre of the triangle. Incentre of a triangle is the intersection point of the angle bisectors of the interior angles of that triangle.
Here, in ΔABC, we can find the incentre of this triangle by drawing the angle bisectors of the interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. Therefore, I is the point equidistant from all the sides of ΔABC.
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संबंधित प्रश्न
In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠A > ∠C and ∠B > ∠D.
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
In a huge park people are concentrated at three points (see the given figure):
A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit.
Where should an ice-cream parlour be set up so that maximum number of persons can approach it?
(Hint: The parlor should be equidistant from A, B and C)
D is a point on the side of the BC of ΔABC. Prove that the perimeter of ΔABC is greater than twice of AD.
ABCD is a trapezium. Prove that:
CD + DA + AB + BC > 2AC.
ABCD is a trapezium. Prove that:
CD + DA + AB > BC.
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that : PN < RN
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that: SN < SR
In ΔPQR, PS ⊥ QR ; prove that: PQ + PR > QR and PQ + QR >2PS.
