Relation Between Two Sides Of A Triangle And Their Opposite Angles
Subina has formed many colourful pieces of cardboard. Now she is trying to make a simple closed figure with three colourful pieces of card board and nails.
She makes –
Measurements of opposite angles equal [ equal/unequal ]
Tapan drew a triangle MAT in which MA=MT. Let’s prove logically and step by step that ΔMAT ∠MAT= ∠MTA
Given: MAT is a triangle where MA = MT. Prove that ∠MAT = ∠MTA.
Construction: In A MAT MX the bisector of ZAMT, is drawn which cuts AT at X.
Proof: ΔMAX and ΔMXT
MA = MT .
∠AMX = ∠TMX [ MX is the bisector of ∠AMT] MX is common to both.
∴ΔMAX = ΔMXT(S-A-S)
∴∠MAX = ∠MTX, i.e., ∠MAT = ∠MTA Proved
Read and Learn More WBBSE Solutions For Class 8 Maths
Now we draw such triangles where measurement of two angles are equal. Let’s find the length of sides of these triangles by scale.
Let measure by scale,
In ΔABC AB=1.5cm, BC=1.5cm, and CA=2.1cm
ΔPQR PQ=1.7cm, QR=1.7cm and RS= 1.7cm
ΔXYZ XY=2cm,YZ=2cm and ZX=3.5cm
It is seen that the length of the sides are equal] [equal/unequal]
Relation Between Two Sides Of A Triangle Exercise
Question 1. Let’s see the isoceles triangles below and let’s write without measuring which angles are equal in measurement in each triangle.
Solution:
In ΔABC ∠BAC = ∠BCA = 70°
In ∠ABC is drawn BD which cuts AC at D.
ΔABD and ΔBCD
∠ABD = ∠CBD = 20° [∵ BD in bisector of Z BAC ]
∠BAD = ∠BCD (Given)
And BD is common.
∴ ΔABD ≅ ΔBCD (A-A-S)
∴ AB = BC
In ΔPQR , ∠QOR = Z PRQ = 45°
∴ PQ = QR
In ΔXYZ ∠YXZ = ∠YZX = 35°
∴ XY = YZ
Question 2. Let’s see the isoceles triangles below and let’s write without measuring which angles are equal in measurement in each triangle.
Solution:
The angles are equal in measurement in each triangle are
In ΔABC AB = BC = 5 cm.
∴∠ACB = ∠BAC
In ΔPQR PQ = PR = 8 cm.
∴∠PRQ = ∠PQR
Question 3. Line segments AB and CD intersect each other at O. Let’s prove that AC and BD are parallel. Let’s write what kind of quadrilateral is ABCD.
Solution:
Given
Line segments AB and CD intersect each other at O.
AB and CD are straight lines which bisect each other at O, i.e.,
AO = OB and CO = OD.
Prove that AC and BD straight lines are parallel to each other.
Proof: In ΔAOD and ΔBOC
AO = OB and CO = OD
And ∠AOD = Vertically Opposite Angle ∠BOC
∴ ΔAOB ≅ ΔBOC (S-A-S)
∴ AD = BC
Now, In ΔAOC and ΔBOD
AO = OB and CO = OD
and AOC = Vertically Opposite Angie ∠BOD
∴ ΔAOC = ΔBOD
∴ AC = BD
The lengths of opposite sides of □ ABCD are equal and the lengths of both diagonals are equal.
ABCD is a rectangle and AC = BD. Proved
Question 4. E and F are two points on two straight lines AB and CD respectively. O is the mid point of line segment EF; we draw a straight line passing through O which intersects AB and CD at P and Q respectively. Let’s prove that O bisects the line segment PQ.
Solution:
Given
E and F are two points on two straight lines AB and CD respectively. O is the mid point of line segment EF; we draw a straight line passing through O which intersects AB and CD at P and Q respectively.
AB//CD and O is the midpoint of, EF i.e., OE = OF Prove that PO = OQ.
Proof: In ΔEOP and ΔQOF
∠EOP = Vertically Opposite Angle ∠FOQ
OE = OF
and ∠PEO = alternate ∠OFQ (∵ AB//CD and EF is a transversal)
∴ ΔEOP ≅ ΔQOF (A – A – S)
∴ PO = OQ, i.e., PQ straight line is bisected at O.
Question 5. If we produce the base of an isoceles triangle in both sides then two exterior angles are formed. Let’s prove that they are equal in measurement.
Solution:
Given
If we produce the base of an isoceles triangle in both sides then two exterior angles are formed.
ABC is an isoceles triangle where AB = AC. Base of isoceles ΔABC base, BC is extended to BD and CE on both sides.
∠ABD = ∠ACE
In ΔABC, AB = AC
∴ ∠ABC = ∠ACB
∵ AB stands on CD straight line
∴ ∠ABC + ∠ABD = 180°
∴ Similarly, ∠ACB + ∠ACE = 180°
∴ ∠ABC + ∠ABD = ∠ACB + ∠ACE
∵ ∠ABC = ∠ACB
∴ ∠ABD = ∠ACE Proved.
Question 6. Let’s prove that the lengths of the medians are equal in an equilateral triangle.
Solution:
Let ABC is an equilateral triangle whose medians are AD, BE, and CF. Prove that AD = BE = CF.
∵ AB = AC (∵ ABC is an equilateral triangle)
∴ 1/2AB=1/2AC
∴ BF = CE
In ΔBCF and ΔBCE,
BF = CE (Proved)
∠FBC = ∠BCE (∵ ABC is an equilateral triangle Each angle is 60°.) and BC is common.
∴ ΔBCF ≅ ΔBCE (S-A-S)
∴ BE = CF
Similarly, it can be proved that AD = BE.
∴ AD = BE = CF. Proved