Advertisements
Advertisements
प्रश्न
The given figure shows a circle with center O. P is mid-point of chord AB.
Show that OP is perpendicular to AB.
Advertisements
उत्तर
Given: in the figure, O is center of the circle, and AB is chord. P is a point on AB such that AP = PB.
We need to prove that, OP ⊥ AB
Construction: Join OA and OB
Proof:
In ΔOAP and ΔOBP
OA = OB ...[Radii of the same circle]
OP = OP ...[Common]
AP = PB ...[Given]
∴ By Side-Side-Side criterion of congruency,
ΔOAP ≅ ΔOBP
The corresponding parts of the congruent triangles are congruent.
∴ ∠OPA = ∠OPB ...[By c.p.c.t]
But ∠OPA + ∠OPB = 180° ...[Linear pair]
∴ ∠OPA = ∠OPB = 90°
Hence, OP ⊥ AB.
APPEARS IN
संबंधित प्रश्न
If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.
In Fig. 10.99, AD ⊥ CD and CB ⊥. CD. If AQ = BP and DP = CQ, prove that ∠DAQ = ∠CBP.
In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.
If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, ∠C = 75°, DE = 2.5 cm, DF = 5cm and ∠D = 75°. Are two triangles congruent?
In the given figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is
In the figure, given below, triangle ABC is right-angled at B. ABPQ and ACRS are squares. 
Prove that:
(i) ΔACQ and ΔASB are congruent.
(ii) CQ = BS.
A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.
In the following figure, AB = EF, BC = DE and ∠B = ∠E = 90°.
Prove that AD = FC.
AD and BC are equal perpendiculars to a line segment AB. If AD and BC are on different sides of AB prove that CD bisects AB.
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
