Advertisements
Advertisements
प्रश्न
In a quadrilateral ABCD, if the perpendicular bisectors of AB and AD meet at P, then prove that BP = DP.
Advertisements
उत्तर

Join A to P.
In Δ AMPand Δ DMP
MP = MP
AM = MD
∠ AMP = ∠ DMP = 90°
Therefore, Δ AMPand Δ DMP are congruent.
DP= AP ....... (i)
In Δ ANP and Δ BNP
NP= NP
AN= NB
∠ANP = ∠BNP = 90°
Therefore, Δ ANP and Δ BNP are congruent.
BP= AP ....... (ii)
From (i) and (ii)
BP= DP
Hence, proved.
APPEARS IN
संबंधित प्रश्न
In each of the given figures; PA = PB and QA = QB.
| i. | ![]() |
| ii. | ![]() |
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.
In triangle LMN, bisectors of interior angles at L and N intersect each other at point A. Prove that:
- Point A is equidistant from all the three sides of the triangle.
- AM bisects angle LMN.
Describe the locus of points at a distance 2 cm from a fixed line.
Describe the locus of a stone dropped from the top of a tower.
Describe the locus of points inside a circle and equidistant from two fixed points on the circumference of the circle.
Describe the locus of a point in rhombus ABCD, so that it is equidistant from
- AB and BC;
- B and D.
Describe the locus of points at distances greater than 4 cm 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.
ΔPBC and ΔQBC are two isosceles triangles on the same base BC but on the opposite sides of line BC. Show that PQ bisects BC at right angles.


