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
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.
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.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
In the figure given below, find a point P on CD equidistant from points A and B.

Describe the locus of points at a distance 2 cm from a fixed line.
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.
In a quadrilateral PQRS, if the bisectors of ∠ SPQ and ∠ PQR meet at O, prove that O is equidistant from PS and QR.
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.
