Updated on: 2026-03-31 | Author: Aarti Kulkarni
= 125°
Proof:
Given: ABCD is cyclic,
∠DCE is the exterior angle of ABCD
To prove: ∠DCE ≅ ∠BAD
ABCD is a cyclic.
By (I) and (II)
∠DCE + ∠BCD = ∠BCD + ∠BAD
∠DCE ≅ ∠BAD
One and only one (unique)
80°
In ▢PQRS,
∠RQT = ∠RSP ...[The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle]
90°
Proof: In given figure,
`{:("AF" = "AE"),("FB" = "BD"),("EC" = "DC"):}}` .....(i) [Tangent Segment theorem]
In ▢ODCE,
`{:(∠"ODC" = 90^circ),(∠"OEC" = 90^circ):}}` ......[Tangent theorem]
Consider R.H.S. = a + b – c
= BC + AC – AB
= (BD + DC) + (AE + EC) – (AF + FB) ......[B–D–C, A–E–C, A–F–B]
= (FB + r) + (AF + r) – (AF + FB) ......[From (i) and (ii)]
= FB + r + AF + r – AF - FB
= 2r
= L.H.S
| Type of arc | Name of the arc | Measure of the arc |
| Minor arc | arc AXB | 100° |
| Major arc | arc AYB | 260° |
72º
Explanation:
ABCD is a cyclic quadrilateral.
Now,
⇒ ∠C = `(2 xx 180º)/5`
Thus, the measure of ∠C is 72º.
Proof: Draw seg GF.
∠EFG = ∠FGH ......[Alternate angles] (I)
∠EFG = `1/2 ("arc EG")` ......[Inscribed angle theorem] (II)
∠FGH = `1/2 ("arc FH")` ......[Inscribed angle theorem] (III)
chord EG ≅ chord FH ........[corresponding chords of congruent arcs]
Chords AC and DE intersect internally at point B.
(i)
Line KL is the tangent to the circle at point L and seg ML is the radius. ......[Given]
In ∆MLK,
(ii)
We know that,
ML = `1/2` MK,
In ∆MLK,
(ii)
chord PQ ≅ chord RS .....[Given]
Now,
m(arc EF) = ∠ECF ...[Definition of measure of arc]
= 90° + 70° ...[From (i) and (ii)]
In a figure, D is the centre of the circle.
ADBC is a quadrilateral.
`{:(∠"QMP" = 90^circ),(∠"QNP" = 90^circ):}}` .....[Tangent theorem]
In ▢MQNP,
Construction: Draw segments XZ and YZ.
Proof:
By theorem of touching circles, points X, Z, Y are collinear.
Now, seg XA ≅ seg XZ ...[Radii of the same circle]
Similarly, seg YB ≅ seg YZ ....[Radii of the same circle]
∠XAZ = ∠ZBY
In given figure, seg CF ⊥ Chord DE
= `1/2 xx 16` .....[Given]
In given figure, AM is the radius of the circle.
In ∆AMN,
Line AB is the tangent to the circle with centre C and radius AC. ......[Given]
Given: AD ⊥ DC
To find: Radius of the circle, i.e., OP.
Solution:
In ▢DSOP,
To find: x
Solution:
= 5 – x ......(iii) [From (i) and given]
Note that seg AB is tangent to the given smaller circle at point C.
∆ABC is an equilateral triangle.
∠CBP = `1/2` ∠ABC ......[Ray BP bisects ∠B]
In ∆ABQ,
∠BAQ = ∠BAC + ∠CAQ .....[Angle addition property]
In ∆CQA,
∠CAQ = ∠CQA .......[From (ii) and (iii)]
∠ABC = `1/2` m (arc ADC) ......[Inscribed angle theorem]
= `1/2 xx 130^circ`
= 65°
Now, ∠CDA = `1/2` m (arc ABC) ......[Inscribed angle theorem]
∠CBE = ∠CDA ......(iiii) [The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle]
chord LM ≅ chord LN
= `2 xx 145^circ/2`
= 145°
In the figure, arc AXC is intercepted by ∠ABC.
In given figure,
∠PQR ≅ ∠PSR ......[Angles inscribed in the same arc are congruent]
= `1/2` m(arc SQ) + `1/2 "m"("arc PR")` .....[Inscribed angle theorem]
= `1/2 ["m"("arc PR") + "m"("arc SQ")]`
Given: line AQ is a tangent.
To find: AP
In the given figure, arc PMQ is a semicircle.
∠MPQ = `1/2` m(arc MQ) .....[Inscribed angle theorem]
To find: ∠CAB
Solution:
In given figure,
Also, ∠ACB = `1/2` m(arc AB) .....[Inscribed angle theorem]
In ∆ACB,
Side PQ ≅ side RQ ...[Given]
But, ∠QPR = `1/2` m(arc QR) .....[Inscribed angle theorem]
In ▢ABCD,
To find: ∠BCD
Solution:
∠BDC = `1/2` m(arc BC) ......[Inscribed angle theorem]
In ∆BCD,
Proof:
seg OA ≅ seg OB ......[Radii of the same circle] (ii)
Chord AB ≅ Chord CD ...(Given)
From (1), (2), and (3), we get,
Chords EN and FS intersect externally at point M.
= `1/2 (125^circ - 37^circ)`
= `1/2 xx 88^circ`
Here, 8 cm < 9 cm
seg OP is the radius of the circle.
Proof:
m∠PQR = `1/2 xx ["m"("arc PTR")]` ......(i) [Inscribed angle theorem]
m∠PSR = `1/2 xx ["m"("arc PTR")]` ......(ii) [Inscribed angle theorem]
Proof: In ∆RMO and ∆RNO,
hypt OR ≅ hypt OR ......[Common side]
seg OM ≅ seg ON ......[Radii of the same circle]
∠MOR ≅ ∠NOR
Similarly ∠MRO ≅ ∠NRO ......[Corresponding angles of congruent triangles]
seg DP and seg DQ are the tangents to the circle.
Given: A is the centre of the circle. Tangents through external point D Touch the circle at the points P and Q.
To Prove: seg DP ≅ seg DQ
Proof:
In ∆PAD and ∆QAD,
seg PA ≅ seg QA .....[Radii of the same circle]
seg AD ≅ seg AD ......[Common side]
Proof: Segment AC is a diameter of the circle.
Arc AXC is intercepted by the inscribed angle ∠ABC
∠ABC = `1/2 "m"("arc AXC")` ......[Inscribed angle theorem]
= `1/2 xx 180^circ` ......[From (i)]
In ΔABC and ΔDBE,
side AB ≅ side DB ......[Radii of the same circle]
side BC ≅ side BE .....[Radii of the same circle]
∠ABC ≅ ∠DBE ......[Measure of congruent arcs]
To find: l(BP) , l(BR)
Solution:
`{:("seg AP" ≅ "seg AQ"),("seg QC" ≅ "seg RC"),("seg BP" ≅ "seg BR"):}}` .......[Tangent Segment theorem]
It is given that two circle having radii 3.5 cm and 4.8 cm touch each other internally.
We know, the distance between the centres of the circles touching internally is equal to the difference of their radii.
Thus, the distance between their centres is 1.3 cm.
Measure of semi circular arc is 180°.