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Home / Geometry / Solution of exercise 8.2| Class Nine and Ten (9 – 10)

# Solution of exercise 8.2| Class Nine and Ten (9 – 10)

PROBLEM – 1. ABCD is quadrilateral inscribed in a circle with centre O. If the diagonals AC and BD intersect at the point E,prove that ∠AOB+∠COD=2∠AER.

General Enunciation: ABCD is quadrilateral inscribed in a circle with centre O. If the diagonals AC and BD intersect at the point E, we have to prove that ∠AOB + ∠COD=2∠AEB.

Particular Enunciation: Given that two chords AC and BD of a circle with centre O, intersect in the interior point E of the circle .Let us join (O, A), (O,B),(O,C) and (O,D). We have to prove that

(∠AOB+ <COD) =2∠AEB.

Proof: ∠AOB and ∠ADB are two angles at the centre and at the circumference respectively on the same are AB.

Again, ∠COD and ∠CAD are two angles at the centre and at the   circumference respectively on the same are CD.

Adding equations (1) and (2) we get

Putting this value in equation (iii) we get, (∠AOB +∠COD)=2∠AEB (proved).

PROBLEM – 2. Two chords AB and CD of the circle ABCD intersect at the point E. show that ∆AED and ∆BEC are equiangular.

General enunciation: Two chords AB and CD of the circle ABCD intersect at the point E. we have two show that AED and BEC are equiangular.

Particular enunciation: Suppose, O is the centre of the circle ABCD. Two chords AB and CD intersect at E. (A,D)and (C,B) are joined. Let us   show that AED and BEC are equiangular.

Proof: ∠DAB and ∠BCD are one some are BD.

∴∠ DAB=∠BCD

Similarly,∠ADC and∠ABC are on the some are AC.

∴ ∠ADC =∠ABC                                                                                                                                                 Now in AED and BEC we get

∠DAE= ∠BCE      [they line on the some arc BC]

∠ADE=∠CBE        [they line on the some arc AC]

And ∠ADE=∠BEC         [Vertically opposite angles]    (Showed)

PROBLEM – 3. In the circle ABCD with the centre O,ADB+BDC=1 right angle. Prove that A ,O and C lie in the some straight line.

General Enunciation: In the circle ABCD with the centre O, ∠ADB+∠BDC=1right angle.

We have to prove that A, O and C line in the some straight line.

Particular Enunciation: Given that in the circle ABCD with the centre O, ∠ADB+ ∠BDC =1 right angle. Let us prove that A, O and C line in some straight line.

Construction: (A, O), (C,O) and (B,O) are joined.

Proof: ∠AOB and ∠ADB are two angles at the centre and at the circumference respectively on the some are AB.

Similarly, ∠BOC and ∠BDC are two angles at the centre and at the circumference respectively on the same BC.

∴ ∠BOC = 2∠BDC ————————– (2)

By adding equations (1) and (2) we get,

∠AOB +∠BOC= 2 ∠ADB = ∠BDC )

∴∠AOB + ∠BOC =2 (right angle) [∠ADB+ ∠BDC= 1 right angle]

=2 (right angles, i.e. 1 straight angle.

Hence, A, O and C are one the same line (Proved)

PROBLEM – 4. In interior of a circle, two chords AB and CD interest at point E. Prove that sum of the angles subtended by the arcs AC and BD at the centre is twice AEC

General Enunciation: Interior of a circle, two chords AB and CD intersect   at a point E. We have to prove that the sum of the angles subtended by the arcs AC and BD at the centre is twice ∠AEC

Particular Enunciation: Given that two chords AB and CD of the circle ABCD with centre O intersect in the interior point E of the circle. Let us join (O, A), (O, B) (O, C) and (O,D).Two arcs AC and BD subtended two angles <AOC and <BOD respectively at the centre. Let us prove that (∠BOD+ ∠AOC) =2∠AEC

Construction: Let as join (B, C)

Proof: ∠AOC and ∠ABC are two angles at the centre and at the circumference respectively on the same arc AC.

∴∠AOC=2∠ABC —————————- (1)

Again, ∠BOD and ∠BCD are two angles at the centre and at the   circumference respectively on the same arc BD

∴ ∠BOD=2∠BCD ————————– (2)

Adding equation (1) and (2) We get,

∠AOC+ <BOD=2(∠ABC+∠BCD) … … (3)

But in BCE, external ∠AEC=∠BCE+∠CBE

∴∠AEC = ∠BCD+∠ABC

Putting this value in equation (3)We get,

(∠AOC+∠BOD) =2∠AEC   (Proved)

PROBLEM – 5. Show that the oblique sides of cyclic trapezium are equal.

General Enunciation: We have show that the oblique sides of a cyclic trapezium are equal.

Particular Enunciation: Suppose, ABCD is the circle with the centre O. ABCD be the trapezium inscribed in that circle. In trapezium ABCD, AB ∥ CD and AD and BC are oblique. Let us show that AD=BC

Construction:   Let us join (O, A),(O,B),(O,C),(O,D) and (B,D).

Proof: ∠BOC and ∠BDC are two angles at the centre and at circumference respectively one some are BD.

∴∠BOC =2∠BDC ————————- (1)

Again, ∠AOD and ABD are two angles at the centre and at circumference respectively on the same are AD

∴∠AOD =2∠ABD ———————————(2)

But AB ∥ CD and BD is there scant.

∴ ∠ABD = ∠BDC

∴  ∠AOD= 2∠BDC ——————————-(3)

∴ From equations (1) and (3) we get ∠BOC =∠AOD

PROBLEM – 6. AB and AC are two chords of a circle; P and Q are the middle points of the two the minor arcs intersected by them. The chord PQ intersects the chords AB and AC at the points D and E respectively. Show that AD =AE.

General Enunciation AB and AC are two chords of a circle; P and Q are the middle points of the two the minor arcs intersected by them. The chord PQ intersects the chords AB and AC at the points D and E respectively. we have to Show that

Particular Enunciation: Suppose, in the circle; ABC with the centre O, AB and AC are two chords. P and Q are the middle points of the two minor arcs intersected by AB and AC respectively. The chord PQ intersects the chords AB and AC at the points d and E respectively. Let us show that AD = AE.

Construction: (A, P) and (P, C) are joined.

Proof: Since, P is the middle points of are AB,

∴ Are AP= are PB

∴ ∠ACP= ∠PAB ———————- (1)

Again Q is the middle point of are AC

∴Arc AQ= arc QC

∴ ∠ACP+∠APQ …. …. (2)

By adding equations (1) and (2) we get,

∠ACP+∠CPQ=∠PAB ∠APQ ————————- (3)

i,e., ∠AED= ∠ACP+∠CPQ —————————— (4)

= ∠ACP+∠CPQ        [From (3)]

=∠AED         [from (4)]

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...