Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
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.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label:
- the locus of the centres of all circles which touch AB and AC,
- the locus of the centres of all the circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC .
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
In Fig. ABCD is a quadrilateral in which AB = BC. E is the point of intersection of the right bisectors of AD and CD. Prove that BE bisects ∠ABC.
Find the locus of points which are equidistant from three non-collinear points.
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.
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?
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect in P. Show that P is equidistant from the opposite sides AB and CD.
