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 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.
Describe the locus of points at a distance 2 cm from a fixed line.
Describe the locus of the centre of a wheel of a bicycle going straight along a level road.
Describe the locus of a stone dropped from the top of a tower.
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.
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.
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.
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.
