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In △PQR, PQ = QR and L, M, N are midpoints of the sides PQ, QP and RP, respectively (Given)
To prove: LN = MN
As two sides of the triangle are equal, so △ PQR is an isosceles triangle
PQ = QR and ∠QPR = ∠QRP ……. (i)
Also, L and M are midpoints of PQ and QR, respectively
PL = LQ = QM = MR = QR/2
Now, consider Δ LPN and Δ MRN,
LP = MR
∠LPN = ∠MRN [From (i)]
∠QPR = ∠LPN and ∠QRP = ∠MRN
PN = NR [N is the midpoint of PR]
By SAS congruence criterion,
Δ LPN ≃ Δ MRN
We know that the corresponding parts of congruent triangles are equal.
So LN = MN
Proved.
Here, AD, BE and CF are medians of △ABC.
Now,
D is the midpoint of BC => BD = DC
Similarly, CE = EA and AF = FB
Since ΔABC is an equilateral triangle
AB = BC = CA …..(i)
BD = DC = CE = EA = AF = FB …………(ii)
And also, ∠ ABC = ∠ BCA = ∠ CAB = 60° ……….(iii)
Consider Δ ABD and Δ BCE
AB = BC [From (i)]
BD = CE [From (ii)]
∠ ABD = ∠ BCE [From (iii)]
By SAS congruence criterion,
Δ ABD ≃ Δ BCE
=> AD = BE ……..(iv)
[Corresponding parts of congruent triangles are equal in measure]
Now, consider Δ BCE and Δ CAF,
BC = CA [From (i)]
∠ BCE = ∠ CAF [From (iii)]
CE = AF [From (ii)]
By SAS congruence criterion,
Δ BCE ≃ Δ CAF
=> BE = CF …………..(v)
[Corresponding parts of congruent triangles are equal]
From (iv) and (v), we have
AD = BE = CF
Median AD = Median BE = Median CF
The medians of an equilateral triangle are equal.
Hence proved
To find: ∠ B and ∠ C.
Here, Δ ABC is an isosceles triangle since AB = AC
∠ B = ∠ C ……… (i)
[Angles opposite to equal sides are equal]
We know that the sum of angles in a triangle = 180°
∠ A + ∠ B + ∠ C = 180°
∠ A + ∠ B + ∠ B= 180° (using (i)
120
0
+ 2∠B = 180
0
2∠B = 180
0
– 120
0
= 60
0
∠ B = 30
o
Therefore, ∠ B = ∠ C = 30
∘
Such that BC ∥ AD and
BC = AD …….(i)
To prove: AB and CD bisect at O.
First, we have to prove that Δ AOD ≅ Δ BOC
∠OCB =∠ODA [AD||BC and CD is transversal]
AD = BC [from (i)]
∠OBC = ∠OAD [AD||BC and AB is transversal]
By ASA Criterion:
Δ AOD ≅ Δ BOC
OA = OB and OD = OC (By c.p.c.t.)
Therefore, AB and CD bisect each other at O.
Hence Proved.
Since AB = AC
=> ∠ABC = ∠ACB ……(i)
[Angles opposite to equal sides are equal]
Since BD and CE are bisectors of ∠ B and ∠ C
∠ ABD = ∠ DBC = ∠ BCE = ECA = ∠B/2 = ∠C/2 …(ii)
Now, Consider Δ EBC = Δ DCB
∠ EBC = ∠ DCB [From (i)]
BC = BC [Common side]
∠ BCE = ∠ CBD [From (ii)]
By ASA congruence criterion, Δ EBC ≅ Δ DCB
Since corresponding parts of congruent triangles are equal.
=> CE = BD
or, BD = CE
Hence proved.
To prove: Both triangles are congruent.
Consider two right triangles such that
∠ B = ∠ E = 90
o
…….(i)
AB = DE …….(ii)
∠ C = ∠ F ……(iii)
Here we have two right triangles, △ ABC and △ DEF
From (i), (ii) and (iii),
By the AAS congruence criterion, we have Δ ABC ≅ Δ DEF
Both triangles are congruent. Hence proved.
From figure,
∠1 = ∠2 [AD is a bisector of ∠ EAC]
∠1 = ∠3 [Corresponding angles]
and ∠2 = ∠4 [alternative angle]
From above, we have ∠3 = ∠4
This implies, AB = AC
Two sides, AB and AC, are equal.
=> Δ ABC is an isosceles triangle.
Since, PQ= PR, so △PQR is an isosceles triangle.
∠ PQR = ∠ PRQ
Now, ∠ PST = ∠ PQR and ∠ PTS = ∠ PRQ
[Corresponding angles as ST parallel to QR]
Since, ∠ PQR = ∠ PRQ
∠ PST = ∠ PTS
In Δ PST,
∠ PST = ∠ PTS
Δ PST is an isosceles triangle.
Therefore, PS = PT.
Hence proved.
Join L and M, M and N, N and L
We have PL = LQ, QM = MR and RN = NP
[Since L, M and N are mid-points of PQ, QR and RP, respectively]
And also, PQ = QR
PL = LQ = QM = MR = PN = LR …….(i)
[ Using mid-point theorem]
MN ∥ PQ and MN = PQ/2
MN = PL = LQ ……(ii)
Similarly, we have
LN ∥ QR and LN = (1/2)QR
LN = QM = MR ……(iii)
From equations (i), (ii) and (iii), we have
PL = LQ = QM = MR = MN = LN
This implies, LN = MN
Hence Proved.
In △BDP and △CDQ
PD = QD (Given)
BD = DC (D is mid-point)
∠BPD = ∠CQD = 90
o
By RHS Criterion: △BDP ≅ △CDQ
BP = CQ … (i) (By CPCT)
In △APD and △AQD
PD = QD (given)
AD = AD (common)
APD = AQD = 90
o
By RHS Criterion: △APD ≅ △AQD
So, PA = QA … (ii) (By CPCT)
Adding (i) and (ii)
BP + PA = CQ + QA
AB = AC
Two sides of the triangle are equal, so ABC is an isosceles.
In Δ BCF and Δ CBE,
∠ BFC = CEB = 90
o
[Given]
BC = CB [Common side]
And CF = BE [Given]
By RHS congruence criterion: ΔBFC ≅ ΔCEB
So, ∠ FBC = ∠ EBC [By CPCT]
=>∠ ABC = ∠ ACB
AC = AB [Opposite sides to equal angles are equal in a triangle]
Two sides of triangle ABC are equal.
Therefore, Δ ABC is isosceles. Hence Proved.
Consider an angle ABC and BP be one of the arms within the angle.
Draw perpendiculars PN and PM on the arms BC and BA.
In Δ BPM and Δ BPN,
∠ BMP = ∠ BNP = 90° [given]
BP = BP [Common side]
MP = NP [given]
By RHS congruence criterion: ΔBPM≅ΔBPN
So, ∠ MBP = ∠ NBP [ By CPCT]
BP is the angular bisector of ∠ABC.
Hence proved
