Let’s see the isosceles triangles given below and without measuring let’s write which two sides are equal in length in each triangle.
It is given that ∆ABC is isosceles and
∠A=70°
∠B=40°
And ∠C=70°
∴ ∠A=∠C
We know angles opposite to the equal sides are equal.
So, BA=BC
Let’s see the isosceles triangles given below and without measuring let’s write which two sides are equal in length in each triangle.
It is given that ∆PQR is isosceles and
∠P=45°
∠R=45°
∴ ∠P=∠R
We know angles opposite to the equal sides are equal.
So, QP=QR
Let’s see the isosceles triangles given below and without measuring let’s write which two sides are equal in length in each triangle.
It is given that ∆XYZ is isosceles and
∠X=35°
∠Z=35°
And ∠Y=110°
∴ ∠X=∠Z
We know angles opposite to the equal sides are equal.
So, XY=YZ
let’s see the isosceles triangles below and without measuring let’s write which angles are equal in measurement in each triangle.
It is given that ∆ABC is isosceles and
AB=5cm
BC=5cm
∴BA=BC
We know angles opposite to the equal sides are equal.
So, ∠A=∠C
let’s see the isosceles triangles below and without measuring let’s write which angles are equal in measurement in each triangle.
It is given that ∆PQR is isosceles and
PQ=8cm
PR=8cm
And QR=7cm
∴PQ=PR
We know angles opposite to the equal sides are equal.
So, ∠Q=∠R
Line segments AB and CD bisect each other at O. Let’s prove that AC and BD are parallel. Let’s write what kind of quadrilateral is ABCD.
Given here two line segments AB and CD bisects each other at O.
∴ OA=OB and OC=OD
Now, in ∆AOC and ∆BOD,
∠AOC=∠BOD (vertically opposite angles)
OC=OD (given)
OA=OB (given)
∴ ∆AOC ≅ ∆BOD (by SAS rule)
∠ACO=∠ODB (by cpct )
∠CAO=∠OBD (by cpct )
But these are also the alternate interior angles.
So, AC||BD
Similarly, ∆AOD ≅ ∆BOC
And AD||BC
Hence ACBD is a parallelogram.
E and F are twp points on two parallel straight-line AB and CD respectively. O is the midpoint of line segment EF. We draw a straight line passing through O which intersect AB and CD at P and Q respectively. Let’s prove that O bisects the line segment PQ.
E and F are two points on straight lines AB and CD respectively.
Also, O is the midpoint of EF and a straight line is drawn passing through O which intersect AB and CD at P and Q respectively.
OE=OF (O is the midpoint)
AB||CD
∠PEO=∠QFO (alternate interior angles) …………… (1)
∠POE=∠QOF (vertically opposite angles) …………. (2)
Now, in ΔPOE and ΔQOF,
OE=OF (O is the midpoint)
∠PEO=∠QFO (from (1))
∠POE=∠QOF (from (2))
∴ ΔPOE ≅ ΔQOF (by ASA rule)
So, OP=OQ (by cpct)
Hence, O bisects PQ
If we produce the base of an isosceles triangle in both sides then two exterior angles formed. Let’s prove that they are equal in measurement.
Let an isosceles ΔEFG in which the side FG is extended on both sides to form two exterior angles.
EF=FG
So, ∠EFG=∠EGF…………(1)
As we know that sum of two interior angles is equal to the exterior angle of a triangle.
∠EFI=∠EFG+∠E…….(2)
∠EGJ=∠EGF+∠E…….(3)
Now, Subtracting equation (2) from (3)
∠EGJ-∠EFI=∠EGF+∠E-∠EFG-∠E
⇒ ∠EGJ-∠EFI=∠EGF -∠EFG=0 (from (1))
⇒ ∠EGJ-∠EFI=0
⇒ ∠EGJ=∠EFI
Proved.
Let’s prove that the length of the medians are equal in length in an equilateral triangle.
Let an equilateral ΔABC with three medians AE, BF and CG
So, E, F and G are the midpoints of BC , AC and AB
∴ BE=EC, AF=FC and AG=GB
AB=BC=AC
⇒ � AB= � BC= � AC
⇒ AG=BE=AF
Now, in ΔAEC and ΔBFA,
EC=AF
AC=BC (given)
∠ACB=∠BAC (each 60° )
∴ ΔAEC ≅ ΔBFA (by SAS rule)
So, AE=BF (by cpct)………(1)
Similarly, ΔCGB≅ ΔBFA (by SAS rule)
So, CG=BF (by cpct)……….(2)
And, ΔCGB≅ ΔAEC
So, CG=AE…………..(3)
From (1),(2) and (3),
AE=BF=CG
Hence, proved.
In trapezium ABCD, AD||BC and ∠ABC=∠BCD, let’s prove that ABCD is an isosceles trapezium.
Given a trapezium ABCD in which AD||BC and ∠ABC=∠BCD
Draw AEꞱBC and DFꞱBC
∠ABC=∠BCD (given)…………..(1)
∠AEB=∠DFC=90°………………(2)
Now, in ΔAEB,
∠BAE+∠AEB+∠ABE=180° (sum of all the angles of a Δ)……….(3)
In ΔDFC,
∠FDC+∠DFC+∠DCF=180° (sum of all the angles of a Δ)……….(4)
Subtracting equation (4) from (3)
∠BAE+∠AEB+∠ABE-∠FDC-∠DFC-∠DCF=180°-180°
⇒ ∠BAE-∠FDC+∠AEB-∠DFC+∠ABE-∠DCF=0
⇒ ∠BAE-∠FDC=0 (from (1) and (2))
⇒ ∠BAE=∠FDC………………..(5)
Now, in ΔBAE and ΔCDF,
AE=DF (by construction)
∠BAE=∠FDC (from (5))
∠AEB=∠DFC (each 90° )
∴ ΔBAE ≅ ΔCDF (by ASA rule)
So, AB=DC (by cpct)
Hence, the trapezium is isosceles as AB=CD.
AB is the hypotenuse of the isosceles right triangle AB. AD is the bisector of ∠BAC and AD intersects BC at D. Let’s prove AC + CD = AB.
Let BC = AC = a and CD = b
In a right-angled triangle BCA,
By Pythagoras theorem,
AB2 = BC2 + AC2
AB2 = a2 + a2
AB = a√2
Given AD = b, we get
DB = BC – CD or DB = a – b
We have to prove that AC + CD = AB
or (a + b) = a√2.
By the angle bisector theorem, we get
C + CD = AB [we know that AC = a, CD = b and AB = a√2]
Hence proved.
Two isosceles triangles ABC and DBC whose AB = AC, DB = DC and they are situated on the opposite side of BC. Let’s prove that AD bisects BC perpendicularly.
Given that ΔABC and ΔDBC are isosceles triangles.
Here, AB=AC and DB=DC
Also, ∠ABC=∠ACB
And ∠DBC=∠DCB
Now, in ΔABD and ΔACD,
AD=AD (common)
AB=AC (given)
BD=DC (given)
∴ ΔABD ≅ ΔACD (by SSS rule)
∠BAD=∠CAD (by cpct)
Again, in ΔABO and ΔACO,
AO=AO (common)
∠BAO=∠CAO (as ∠BAD=∠CAD)
AB=AC (given)
∴ ΔABD ≅ ΔACD (by SAS rule)
So, BO=OC (by cpct)
And ∠AOB=∠AOC(by cpct)
Also, ∠AOB+∠AOC=180° (straight angle)
⇒ ∠AOB+∠AOB=180°
⇒ 2∠AOB=180°
⇒ ∠AOB=90°
So, ∠AOC=∠AOB=90°
Hence, AD bisects BC perpendicularly.
Two line segments PQ and RS intersect each other at X in such a way that XP=XR and ∠PSX =∠ROX. Let’s prove thatPXS ROX.
Given that two Two line segments PQ and RS intersect each other at X in such a way that XP=XR and ∠PSX =∠RQX
Now, ∠PSX=∠RQX and ∠RXQ=∠PXS (vertically opposite angles)……………(1)
In ΔRXQ,
∠RQX+∠RXQ+∠XRQ=180° ……………(2)
And in ΔPXS,
∠PXS+∠PSX+∠XPS=180° ………………(3)
Subtracting equation (2) from (3),
∠RQX+∠RXQ+∠XRQ-∠PXS-∠PSX-∠XPS=180° -180°
⇒ ∠RQX-∠PXS +∠RXQ-∠PSX +∠XRQ -∠XPS=0
⇒ ∠XRQ -∠XPS=0 (from (1))
⇒ ∠XRQ =∠XPS ……………..(4)
Now, in ΔPSX and ΔRQX,
XR=XP (given)
∠XRQ =∠XPS (from (4))
∠RXQ=∠PXS (vertically opposite angles)
∴ ΔPSX ≅ ΔRQX (by ASA rule)