#### Solved Examples and Worksheet for Properties of Isosceles Triangles

Q1In the figure, MN || PR, LBN = x°, AB¯ = BC¯. Find mABC if x = 63.

A. 52
B. 58
C. 63
D. 54

Step: 1
mBAC = mLBN
[Corresponding angles are congruent.]
Step: 2
mBAC = 63
[Substitute mLBN = 63.]
Step: 3
In ΔABC, AB¯ = BC¯ BCA = BAC
[Isosceles triangle theorem.]
Step: 4
mBCA = 63
[Step 2.]
Step: 5
mABC = 180 - (mBAC + mBCA)
[Triangle Angle-Sum theorem.]
Step: 6
mABC = 180 - (63 + 63)
[Substitute.]
Step: 7
mABC = 54
[Simplify.]
Q2In the figure, PQ¯ = PR¯. mCPR = x. If x = 108, find mPQR.

A. 82
B. 108
C. 118
D. 72

Step: 1
CD || AB
[mDCA = mBAC = 90.]
Step: 2
mPRQ + mCPR = 180
[Same side Interior angles theorem.]
Step: 3
mPRQ + 108 = 180
[Substitute mCPR = 108.]
Step: 4
mPRQ = 72
[Simplify.]
Step: 5
In ΔPQR, PQ¯ = PR¯ PQR = PRQ
[Isosceles triangle theorem.]
Step: 6
mPQR = 72
[Substitute mPRQ = 72.]
Q3In the figure, AB = BX and AC = CY. mABC = a and mACB = b. What is the measure of XAY ? [Given a = 63 and b = 63.]

A. 113
B. 117
C. 121
D. 119

Step: 1
In ΔXAB, ABC = BXA + BAX
[In a triangle, an exterior angle is equal to the sum of the opposite interior angles.]

Step: 2
In ΔXAB, AB = BX BXA = BAX
[Isosceles triangle theorem.]
Step: 3
mABC = 2(mBAX)
[Step 1 and Step 2.]
Step: 4
mBAX = 12(mABC)
[Rearrange.]
Step: 5
mBAX = 12(63)
[Substitute.]
Step: 6
mBAX = 31.5
[Simplify.]
Step: 7
Similarly, in ΔACY, mCAY = 12(mACB)
Step: 8
mCAY = 12(63)
[Substitute.]
Step: 9
mCAY = 31.5
[Simplify.]
Step: 10
In ΔABC, mBAC = 180 - (mABC + mACB)
[Triangle-Angle-Sum theorem.]
Step: 11
mBAC = 180 - (63 + 63)
[Substitute.]
Step: 12
mBAC = 54
[Simplify.]
Step: 13
mXAY = mXAB + mBAC + mCAY
Step: 14
Here a = 63, b = 63, e = 54, c = 31.5 and d = 31.5.
Step: 15
mXAY = 31.5 + 54 + 31.5
[Substitute.]
Step: 16
mXAY = 117
Q4OB¯ and OC¯ are the bisectors of B and C. If mA = 83, then find mBOC.

A. 131.50
B. 121.50
C. 129.50
D. 136.50

Step: 1
In ΔABC, mA + mB + mC = 180
[Triangle Angle-Sum theorem.]
Step: 2
In ΔABC, AB = AC B = C
[Isosceles triangle theorem.]
Step: 3
83 + mB + mB = 180
[Step 1 and step 2.]
Step: 4
2(mB) = 97
[Simplify.]
Step: 5
mB = 48.50
[Solve for B.]
Step: 6
mC = mB = 48.50
[Step 2.]
Step: 7
mOBC = 12(mB) = 12(48.50) = 24.25
[OB¯ is the bisector of B.]
Step: 8
mOCB = 12(mC) = 12(48.50) = 24.25
[OC¯ is the bisector of C.]
Step: 9
In ΔBOC, mBOC = 180 - (mOBC + mOCB)
[Triangle Angle sum theorem.]
Step: 10
mBOC = 180 - (24.25 + 24.25)
[Substitute.]
Step: 11
mBOC = 131.50
[Simplify.]
Q5In the figure, AP¯ is drawn parallel to BC¯. Also, AP¯ bisects CAD. If mB = 57, then find mC.

A. 66
B. 57
C. 62
D. 61

Step: 1
AP¯ || BC¯
[Given.]
Step: 2
Let mDAP be x.
Step: 3
mB = mDAP
[Corresponding angles postulate.]
Step: 4
mPAC = mDAP
Step: 5
mPAC = x
[Substitute mDAP = x.]
Step: 6
mACB = mPAC
[Alternate interior angles theorem.]
Step: 7
mC = mDAP
[mACB = mC and step 4.]
Step: 8
mB = mC
[Step 3 and Step 7.]
Step: 9
mC = 57
[mB = 57.]
Q6ABC is a triangle. The bisector of the BCA meets AB¯ at X. A point Y lies on CX¯ such that AX¯ = AY¯. If mCAY = 36, then find mABC.
A. 45
B. 60
C. 90
D. 36

Step: 1
Let mAXY be α and mXCB be θ.
Step: 2

Step: 3
mXCA = mXCB = θ
[CX¯ is bisector of BCA.]
Step: 4
In ΔAXY, AX = AY mAYX = mAXY
[Isosceles triangle theorem.]
Step: 5
α = (mCAY) + (θ)
Step: 6
mCAY = (α - θ)
[Rearrange.]
Step: 7
In ΔXBC, mAXC = mXBC + mXCB
[Exterior angle theorem.]
Step: 8
α = (mXBC) + θ
[Substitute.]
Step: 9
mXBC = (α - θ)
[Rearrange.]
Step: 10
mXBC = mABC
[Same angle.]
Step: 11
mXBC = mCAY
[Step 6 and step 11.]
Step: 12
mXBC = mCAY = mABC = 36
[Given that mCAY = 36.]
Q7In ΔABC, AB¯ = AC¯. If BD¯ and CE¯ are the medians of ΔABC and BD = 8 cm, then what is the length of CE?

A. 10 cm
B. 8 cm
C. 6 cm
D. 13 cm

Step: 1
As AB¯ = AC¯ B = C.
[Isosceles Triangle.]
Step: 2
ΔEBC = ΔDCB
[SAS property.]
Step: 3
CE = BD = 8 cm
[ΔEBC ΔDCB.]
Correct Answer is :    8 cm
Q8In the figure, AB = AD = CD. Find the value of x in degrees.

A. 70°
B. 55°
C. 45°
D. 35°

Step: 1
[Isosceles triangle theorem.]
Step: 2
[Exterior angle theorem.]
Step: 3
[Isosceles triangle theorem.]

Step: 4
[Triangle-Angle-Sum theorem.]
Step: 5
2x + 2x + 40° = 180°
Step: 6
4x = 180° - 40° = 140°
Step: 7
Therefore, x = 140°4 = 35°
Q9In the figure shown, AB = AC and CE = DE. Find the value of x in degrees.

A. 65°
B. 55°
C. 70°
D. 60°

Step: 1
Let m∠ABC be y.
Step: 2
In ΔABC, AB = AC ⇒ m∠ACB = m∠ABC = y
[Isosceles triangle theorem.]
Step: 3
m∠DCE = m∠ACB = y
[Vertically Opposite angles are equal.]
Step: 4
In ΔCDE, CE = DE ⇒ m∠CDE = m∠DCE = y
[Isosceles triangle theorem.]

Step: 5
In ΔABC, m∠ABC + m∠ACB + m∠BAC = 180°
[Triangle-Angle-Sum theorem.]
Step: 6
y + y + 70° = 180°
Step: 7
2y = 180° - 70° = 110°
Step: 8
So y = 110°2 = 55°
Step: 9
in ΔCDE, m∠DCE + m∠CDE + m∠CED = 180°
[Triangle-Angle-Sum theorem.]
Step: 10
y + y + x = 180°
Step: 11
x = 180° - 2y = 180° - 2 × 55°
Step: 12
Therefore, x = 180° - 110° = 70°
Q10In ΔABC, BD¯ and CE¯ are altitudes and BD = CE. Then which of the following statements is incorrect ?

A. ΔABC may be an equilateral triangle
B. ΔABC may be an isoceles triangle
C. ΔABC is a scalene triangle
D. ΔABC is an isoceles or an equilateral triangle

Step: 1

Step: 2
Area of ΔABC = 12 × AC × BD = 12 × AB × CE
[Area of triangle = 12× base × altitude.]
Step: 3
So, AB = AC
[BD = CE.]
Step: 4
Therefore, ΔABC is an isosceles or an equilateral triangle
[From Step 3]
Correct Answer is :   ΔABC is a scalene triangle