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Practice each question, then open the answer or worked solution.
Find all the angles of an equilateral triangle.
In equilateral triangle,
All the sides are equal.
Therefore, all angles are also equal
Let the angles of an equilateral triangle = x
According to angle sum property,
We know that the sum of the interior angles is equal to 180o.
x+x+x=180o
3x=180
x=60°
Therefore, all the angles of an equilateral triangle are 60°
The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown in Fig. 7.12. Prove that the image is as far behind the mirror as the object is in front of the mirror.
[Hint: CN is normal to the mirror. Also, angle of incidence = angle of reflection].
Let AB intersect LM at O. We have to prove that AO = BO.
Now, ∠i = ∠r …(1)
∠B = ∠i [Corresponding angles] …(2)
And ∠A = ∠r [Alternate interior angles] …(3)
From (1), (2) and (3), we get
∠B = ∠A
⇒ ∠BCO = ∠ACO
In ΔBOC and ΔAOC we have
∠1 = ∠2 [Each = 90o]
OC = OC [Common side]
And ∠BCO = ∠ACO [Proved above]
ΔBOC ≅ ΔAOC [ASA congruence rule]
Hence, AO = BO [CPCT]
ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Fig. 7.13). To prove that ∠BAD = ∠CAD, a student proceeded as follows:
In Δ ABD and Δ ACD,
AB = AC (Given)
∠B = ∠C (because AB = AC)
And ∠ADB = ∠ADC
Therefore, Δ ABD Δ Δ ACD (AAS)
So, ∠BAD = ∠CAD (CPCT)
What is the defect in the above arguments?
[Hint: Recall how ∠B = ∠C is proved when AB = AC].
In Δ ABD and Δ ADC, we have
∠ADB = ∠ADC
According to the question,
AB = BC
AD = AD [Common side]
By RHS criterion of congruence,
We have,
Δ ABD ≅ Δ ACD
∠BAD = ∠CAD [CPCT]
Hence Proved.
P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meet BC at Q, prove that BPQ is an isosceles triangle.
To prove: BPQ is an isosceles triangle.
According to the question,
Since, BP is the bisector of ∠ABC,
∠1 = ∠2 … (1)
Now, PQ is parallel to BA and BP cuts them
∠1 = ∠3 [Alternate angles] … (2)
From equations, (1) and (2),
We get
∠2 = ∠3
In Δ BPQ,
We have
∠2 = ∠3
PQ = BQ
Hence, BPQ is an isosceles triangle.
ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ABC and ADC.
According to the question,
In ΔABC and ΔCBD,
We have
AB = BC
AD = CD
BD = BD [Common side]
ΔABC ≅ ΔCBD [By SSS congruence rule]
⇒ ∠1 = ∠2 [CPCT]
And ∠3 = ∠4
Hence, BD bisects both ∠ABC and ∠ADC.
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2 AD.
Given: A right angles triangle with AB = AC bisector of ∠A meets BC at D.
To prove: BC = 2AD
Proof:
According to the question,
In right Δ ABC,
AB = AC
Since, hypotenuse is the longest side,
BC is hypotenuse
∠BAC = 90o
Now,
In Δ CAD and Δ BAD,
We have,
AC = AB
Since, AD is the bisector of ∠A,
∠1 = ∠2
AD = AD [Common side]
Now,
By SAS criterion of congruence,
We get,
Δ CAD ≅ Δ BAD
CD = BD [CPCT]
Since, Mid-point of hypotenuse of a right triangle is equidistant from the 3 vertices of a triangle.
AB = BD = CD …(1)
Now, BC = BD + CD
⇒ BC = AD + AD [Using eq.(1)]
⇒ BC = 2AD
Hence, proved.
O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that Δ OCD is an isosceles triangle.
According to the question,
It is given that: A square ABCD and OA = OB = AB.
To prove: Δ OCD is an isosceles triangle.
Proof:
In square ABCD,
Since ∠1 and ∠2 is equal to 90o
∠1 = ∠2 …(1)
Now, in Δ OAB , we have
Since ∠3 and ∠4 is equal to 60o
∠3 = ∠4 …(2)
Subtracting equations (2) from (1),
We get
∠1−∠3 = ∠2 −∠4
⇒ ∠5 = ∠6
Now,
In Δ DAO and Δ CBO,
AD = BC [Given]
∠5 = ∠6 [Proved above]
OA = OB [Given]
By SAS criterion of congruence,
We have
Δ DAO ≅ Δ CBO
OD = OC
⇒ Δ OCD is an isosceles triangle.
Hence, proved.
ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.
Given: Δ ABC and Δ DBC on the same base BC. Also, AB = AC and BD = DC.
To prove: AD is the perpendicular bisector of BC i.e., OB = OC
Proof: In Δ BAD and Δ CADwe have
AB = AC [Given]
BD = CD [Given]
AD = AD [common side]
So, by SSS criterion of congruence, we have
Δ BAD ≅ Δ CAD
∠1 = ∠2 [CPCT]
Now, in Δ BAO and Δ CAO, we have
AB = AC [Given]
∠1 = ∠2 [Proved above]
AO = AO [Common side]
So, by SAS criterion of congruence, we have
Δ BAO ≅ Δ CAO
BO = CO [CPCT]
And, ∠3 = ∠4 [CPCT]
But, ∠3+∠4 =180o [Linear pair axiom]
⇒ ∠3+∠3 =180
⇒ 2∠3 =180
⇒ ∠3 =180/2
⇒ ∠3 = 90o
Since BO = CO and ∠3 = 90o,
AD is perpendicular bisector of BC.
Hence, proved.
ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD.
According to the question,
In Δ ADC and Δ BEC,
We have
AC = BC [Given] …(1)
Since ∠ADC and ∠BEC = 90o
∠ADC = ∠BEC
∠ACD = ∠BCE [Common angle]
Δ ADC ≅ Δ BEC [By ASA congruence rule]
CE = CD … (2) [CPCT]
Subtracting equation (2) from (1),
We get
AC – CE = BC – CD
⇒ AE = BD
Hence, proved.
Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.
According to the question,
We have, Δ ABC with median AD.
To prove:
AB + AC > 2AD
AB + BC > 2AD
BC + AC > 2AD
Construction:
Extend AD to E such that DE = AD
Join EC.
Proof:
In Δ ADB and Δ EDC ,
AD = ED [By construction]
∠1 = ∠2 [Vertically opposite angles are equal]
DB = DC [Given]
So, by SAS criterion of congruence, we have
Δ ADB ≅ Δ EDC
AB = EC [CPCT]
And ∠3 = ∠4 [CPCT]
Now, in Δ AEC,
Since sum of the lengths of any two sides of a triangle must be greater than the third side,
We have
AC + CE > AE
⇒ AC + CE > AD + DE
⇒ AC +CE > AD+ AD [\AD = DE]
⇒ AC + CE > 2AD
⇒ AC + AB > 2AD [∵AB = CE]
Similarly,
We get,
AB + BC > 2AD and BC + AC >2AD.
Hence, proved.