Question

Prove with proper explanation!!!

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Answer 1

1)given

ot bisect angle PTR

so

angle RTO=angle OTP

now in triangle TRO and triangle OTP

ro=op(radius of circle is equal)

angle RTO=angle OTP(given explained above too)

to=to(common)

triangle TRO and OTP are congruent

so pt=rt by (cpct)

2 no not explain well in screen so use notebook need drawing too

hope it helps you

Answer 2

ANSWER :

GIVEN :

in circle at centre O

QP and RS are chords such that

angle OTP =angle OTR

TO PROVE :

PT=RT

ST=TQ

CONSTRUCTION :

draw OA perpendicular to QP and OB perpendicular to RS

PROOF :

In ∆OAT and ∆OBT

angle OAT = angle OBT......(Each 90°(by construction))

also

angle OTA= angle OTB........(Given)

also

OT=OT......(common side)

hence by AAS congruence criteria

∆OTA is congruent to ∆OTB

so by CPCT

OA=OB

AT=BT......1

Now in circles

by the theorem that if two chords are equidistant from centre of a circle then they are equal..

as OA = OB

=> PQ=RS

now dividing both sides by 2

PQ/2 = RS/2.....W

but as OA and OB Are drawn as perpendicular by the property that perpendicular drawn on chord bisect the chord

we have AP = AQ=PQ/2

AND BR=BS=RS/2

So from 2

PQ/2=RS/2

AP = BR

Adding AT on both sides

AP+AT=BR+AT

BUT FROM 1

AT= BT

=> AP+AT =BR+BT

=> PT=RT........3

Now

PT = PQ-TQ

AND RT = RS-TS

by 3

PT=RT

PQ-TQ=RS-TS

PQ-RS=TQ-TS

but PQ=RS

So

TQ-TS=0

TQ=TS......4

hence from 3 and 4

PT=RT

ST=TQ

HENCE PROVED