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

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
RELATED QUESTIONS
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.
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point F.
Prove that:

F is equidistant from AB and AC.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Draw an ∠ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC; and also equidistant from A and B.
The speed of sound is 332 metres per second. A gun is fired. Describe the locus of all the people on the earth’s surface, who hear the sound exactly one second later.
Describe the locus of points at distances greater than or equal to 35 mm from a given point.
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
Find the locus of points which are equidistant from three non-collinear points.
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?


