English

ΔPbc and δQbc Are Two Isisceles Triangles on the Same Base. Show that the Line Pq is Bisector of Bc and is Perpendicular to Bc.

Advertisements
Advertisements

Question

ΔPBC and ΔQBC are two isosceles triangles on the same base. Show that the line PQ is bisector of BC and is perpendicular to BC.

Sum
Advertisements

Solution

Given: ΔPBC and ΔQBC are two isosceles triangles on the same base BC.
To prove: Line PQ is the perpendicular bisector of BC.
Proof: In ΔPBC, PB = PC
Since, the locus of a point equidistant from B and C is the perpendicular bisector of 1 of the line segment BC

∴ P lies on 1
Similarly Q lies on 1
Therefore, PQ is the perpendicular bisector of BC.
Hence proved.

shaalaa.com
  Is there an error in this question or solution?
Chapter 17: Loci - Figure Based Questions

APPEARS IN

R.S. Aggarwal Mathematics [English] Class 10 ICSE
Chapter 17 Loci
Figure Based Questions | Q 10

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Construct a right angled triangle PQR, in which ∠Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR. 


In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC. 


Describe the locus of points at a distance 2 cm from a fixed line. 


Describe the locus of a point in rhombus ABCD, so that it is equidistant from

  1. AB and BC;
  2. B and D.

Describe the locus of points at distances greater than or equal to 35 mm from a given point. 


In the given figure, obtain all the points equidistant from lines m and n; and 2.5 cm from O. 


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. 


In Fig. AB = AC, BD and CE are the bisectors of ∠ABC and ∠ACB respectively such that BD and CE intersect each other at O. AO produced meets BC at F. Prove that AF is the right bisector of BC.


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.


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×