Question

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Solve perfectly and clearly with proper explanations....✌

No. 10 and 12 not no. 11​

Answer 1

10 .Given ,

    o =centre

    CM⊥OA

     CN⊥OB

      TO PROVE AC=CB

in triangle AMC and BNC  ,

∠AMC=∠BNC{given perpendicular}

MC=CN {GIVEN}

so AC=CB{ lines opposite to equal angles are equal}

hence proved

12. given ∠APB=∠CQD

TO PROVE  arcAB=arcCD

  CONSTRUCTION:

join P,Q,B&C TO O

  from this QO=OP=OM=OC{RADIUS}

Taking triangle POB &QOC

 OP=OQ{radius}

OB=OC{radius}

∠POB=∠QOC{vertically opposite angles }

hence by SAS property ΔPOB≅ΔQOC

SO CQ=PB by  CPCT

Since in sector APB & DQC

angles are equal and PB=QC

         so arcAB =arcCD  ( since arc opposite to equal angles are equal-only if one angle and side are equal

hence proved

Answer 2

10 )

To proof :-

C is the midpoint of the arc AB i.e arc AC = arc BC.

Construction :-

• Join OC

Given :-

• CM ⊥ OA
• CN ⊥ OB
• CM = CN

Solution :-

To proof both the ∆ congruent

∆OMC and ∆ONC :-

1. OC = common

2. /_ OMC = /_ ONC [•.• Both are 90 ° ]

3. MC = NC

Therefore, both the traingles ∆ OMC are congruent to ∆ ONC . ( Proved by SAS criterion)

OM = ON ( CPCTC)

∆ MAC & ∆ MCB :-

1. CM = CN

2. /_CMA = /_ CNB [ Both are 90 ° since 180 ° - /_ OMC ( 90° ) = /_CMA (90 ° and 180 ° - /_ONC (90°) = CNB ( 90 °) ]

3. MA = NB [ Since OA = OB ( radius of the circle is equal ) OM = ON (By CPCTC proved from the first congruency of the triangles)

Therefore , ∆ MAC congruent to ∆NCB [ by RHS criterion ]

AC = BC or C is the mid point of AB segment ( CPCTC)

Therefore, C is the mid point of the arc AB

since by angle properties theorem if C is the mid point of the AB segment it is also the mid point of the AB arc.

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