Please give me answer
Answers
Step-by-step explanation:
Given :-
AB is a line segment.P and Q are the opposite points of AB such that each of them equidistant from the points A and B .
Required To Proof :-
Show that the line PQ is perpendicular bisectors of AB.
Proof :-
Given that
AB is a line segment.P and Q are the opposite points of AB such that each of them equidistant from the points A and B .
PA = PB and AQ = AB
In ∆ PAQ and ∆ PBQ
PA = PB (Given )
AQ = AB (Given )
PQ = PQ ( Common side )
By SSS property
∆ PAQ =~ ∆ PBQ
Since , AP = PB => < PAB = < PBA
Since ,the angles opposite to equal sides are equal.
AQ = AB => < QAB = < QBA
Since ,the angles opposite to equal sides are equal.
So,
∆ APB and ∆ AQB are the Isosceles triangles.
From ∆ PAC and ∆PBC
PA = PB
< PAB = < PBA
PC = PC ( Common side)
By SAS Property ∆ PAC =~ ∆ PBC
=> AC = CB and < PCA = < PBA -----(1)
=> C is the mid point of AB
Since The corresponding parts in the congruent triangles are equal.
AB is a linesegment then
< ACP + <BCP = 180°
=> < ACP + < ACP = 180° (from (1))
=> 2 < ACP = 180°
=> < ACP = 180°/2
=> < ACP = 90°
and
< BCP = 90°
We have ,
C is the mid point of AB
and
< ACP = 90° and < BCP = 90°
=> PC is the perpendicular and PC is the bisector
=> PQ is the perpendicular bisector of AB
Hence, Proved.
Used formulae:-
→ The angles opposite to equal sides are equal.
→ The corresponding parts in the congruent triangles are equal.
→ The sum of angles on a line is 180°
Step-by-step explanation:
in figure ab is a straight line on which pb and ap sides are standing.
So, in triangle aop and bop
op is a bisector of angle apb and ab line. then, angle apo-bpo (op is the bisector of angle apb)
ao ob (op is perpendicular and bisector of ab) and, angle aop bop=90 degree (op
perpendicular ab making right angle) Or,op op (common) then triangle aop congruent to triangles bop.
hence proved that op is that bisector and parpendicular of AB. hence proved....
hoped helps..
