9B AFL 5th OCT 2020. by Lakshmi Sathyanarayanan

Syed Zidane Imran Mushtaq 9B

2020-10-05 08:54:34 UTC

Given:-
l is the perpendicular bisector of AB
P lies on l

To Prove:
P is equidistant from A and B

Proof:
PC = PC (Common Side)
AC = AC (l is the perpendicular bisector of AB)

Hence, by SAS criteria of congruency:
Triangle PCA is congruent to Triangle PCB

Hence by CPCT,
AP = PB

Hence, P is equidistant from A and B.

Hence proved.

Q.E.D
◼

Joshua James Proof:

joshuadaboss360 — 2020-10-05 08:54:58 UTC

AB = BC ; In ∆ACP & ∆PCB
AC = CB (GIVEN)
PQ=PQ (common)
∴ by SAS cong. ∆ACP ≅ ∆PCB
∴ by CPCT PA = PB

2020-10-05 08:55:21 UTC

Given:- l is the perpendicular

i is the perpendicular bisector of triangle

2020-10-05 08:55:39 UTC

Krishaanth Sairajaganesh 9B

2020-10-05 08:55:53 UTC

Given:-
AB has a perpendicular bisector l
and
point p lies on l

To prove:-
P is equidistant from A and B

Proof:-
PC = PC (Common)
AC=BC (l is the perpendicular bisector of AB)

∴ by SAS criteria ∆PAC ≅ ∆PBC

∴ by CPCT, AP =BP

Therefore because AP = BP, P is equidistant from A and B. (Proved)

l is the perpendicular bisector of AB.

2020-10-05 08:55:58 UTC

P lies on l
TO Prove : -
P is equidistant from A and B
Proof : -
Angle PCB = Angle PCA = 90°
AC = BC
PC = PC
∴ BY SAS Congurency :
Triangle PCA is congurent to Triangle PCB
∴ BY CPCT : PA = PB
Therfore,
P is equidistant from A and B

Girjesh Subbarayalu

Shaurya

2020-10-05 08:56:20 UTC

I is the perpendicular bisector of triangle

AB = BC ; < ACP = < BCP = 90°

B and A are equidistant from P

Angle PCB and Angle PCA = 90 degrees

pc = pc = 90 degree

i is the perpendicular bisector of triangle apb p = equidistant because ab = bc

2020-10-05 08:56:22 UTC

so triangle acp = bcp = 90
pc = pc
so ac = ac

so by using the side angle side congruency p is equidistant

so by cpct
ap = bp

hence,proved

s.vrishnviswa

Fateen Ajaz | 9B

2020-10-05 08:58:13 UTC

I is the bisector of AB
P also lies on I

Prove: A and B are equidistant from P
Proof: Angle PCA and Angle PCB are 90 degrees.
PC = PC (since they are common sides)
AC = AB (I is ⊥ bisector to AB)
Angle PCB = Angle PCA (90 degrees)

Because of SAS, PAC is congruent to PBC

∴ By CPCT AB = BP

Hence, A and B are equidistant from P

Raza Mohamed

2020-10-05 08:58:31 UTC

I is the perpendicular bisector of triangle

AB = BC ; Prove: A and B are equidistant from P
Proof: Angle PCA and Angle PCB are both 90 degrees
PC = PC
so by SAS Congruency is equidistant

therefore by cpct
ap=bp
proved

JASWANT ab is perpendicular bisector of ab .

2020-10-05 08:58:57 UTC

angle p is equidistant from a and b
to prove: angle PCA = angle PCA =90 degree
PC=PC - common
AC=AB- l prependicular to bisector of ab
hence SAS congruency
tri PCA is congrent to tri PCB

2020-10-05 08:59:07 UTC

I is the perpendicular bisector of triangle AB = BC ;

jashan

2020-10-05 08:59:50 UTC

i is the perpendicular bisector

Samarth

2020-10-05 09:00:09 UTC

AB=BC
angle ACP=angle BCP=90

Thanmay Prebeesh

2020-10-05 09:01:36 UTC

Given,
I is the perpendicular bisecector of the line AB and point p lies on l

To prove:-
P is equidistant from A and B

Proof:-

AC=BC (l is the perpendicular bisector of AB)
PC = PC (Common)

Triangle PAC = Triangle PBC
(SAS Congruency)

AP =BP (CPCT)

Therefore because AP = BP, P is equidistant from A and B.

HENCE, PROVED

2020-10-05 09:06:14 UTC

george jacob

2020-10-05 09:06:50 UTC

Shoaib

2020-10-05 09:07:18 UTC

AC=BC

In triangle PCA And triangle PCB

AC=BC

angle PCA= angle PCB =90 degrees

(l is the perpendicular on AB)

PC=PC

therefore triangle PCA ≅ triangle PCB

therefore PA=PB henceproved