Advertisements
Advertisements
प्रश्न
In Δ ABC, the perpendicular bisector of AB and AC meet at 0. Prove that O is equidistant from the three vertices. Also, prove that if M is the mid-point of BC then OM meets BC at right angles.
Advertisements
उत्तर
Since O lies on the perpendirular bisector of AB, O is equidistant from A and B.
OA = OB ........ (i)
Again, O lies on the perpendirular bisector of AC, O is equidistant from A and C.
OA = OC ......... (ii)
From (i) and (ii)
OB= OC
Now in Δ OBM and Δ OCM,
OB = OC (proved)
OM=OM
BM =CM ( M is mid-point of BC)
Therefore, Δ OBM and Δ OCM are congruent.
∠ OMB= ∠ OMC
But BMC is a straight line, so
∠ OMB =∠ OMC = 90°
Thus, OM meets BC at right angles.
APPEARS IN
संबंधित प्रश्न
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
Use ruler and compasses only for this question.
- Construct ΔABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
- Construct the locus of points inside the triangle which are equidistant from BA and BC.
- Construct the locus of points inside the triangle which are equidistant from B and C.
- Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Describe the locus of the centres of all circles passing through two fixed points.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
In the given figure, obtain all the points equidistant from lines m and n; and 2.5 cm from O.
Show that the locus of the centres of all circles passing through two given points A and B, is the perpendicular bisector of the line segment AB.
ΔPBC, ΔQBC and ΔRBC are three isosceles triangles on the same base BC. Show that P, Q and R are collinear.
Given: ∠BAC, a line intersects the arms of ∠BAC in P and Q. How will you locate a point on line segment PQ, which is equidistant from AB and AC? Does such a point always exist?
Use ruler and compasses for the following question taking a scale of 10 m = 1 cm. A park in a city is bounded by straight fences AB, BC, CD and DA. Given that AB = 50 m, BC = 63 m, ∠ABC = 75°. D is a point equidistant from the fences AB and BC. If ∠BAD = 90°, construct the outline of the park ABCD. Also locate a point P on the line BD for the flag post which is equidistant from the corners of the park A and B.
