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2 answers:
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The missing figure is attached
(i) m∠APB = 90° ⇒ proved
(ii) AD = DP and PB = PC = BC ⇒ proved
(iii) DC = 2 AD ⇒ proved
Step-by-step explanation:
ABCD is a parallelogram in which:
- m∠A = 60°
- The bisectors of ∠A and ∠B meet DC at P
In parallelogram ABCD:
∵ m∠A = m∠C ⇒ opposite angles
∵ m∠A = 60°
∴ m∠C = 60°
∵ m∠A + m∠B = 180 ⇒ two adjacent supplementary angles
∴ 60 + m∠B = 180 ⇒ subtract 60 from both sides
∴ m∠B = 120°
∵ m∠B = m∠D ⇒ opposite angles
∴ m∠D = 120°
∵ AP is the bisector of angle A
- That means AP divide ∠A into two equal parts
∴ m∠BAP = m∠DAP = m∠A
∴ m∠BAP = m∠DAP = (60°)
∴ m∠BAP = m∠DAP = 30°
∵ BP is the bisector of angle B
- That means BP divide ∠B into two equal parts
∴ m∠ABP = m∠CBP = m∠B
∴ m∠ABP = m∠CBP = (120°)
∴ m∠ABP = m∠CBP = 60°
(i)
In ΔAPB
∵ m∠BAP = 30° ⇒ proved
∵ m∠ABP = 60° ⇒ proved
- Sum of the measures of the interior angles in a Δ is 180°
∴ m∠APB + m∠BAP + m∠ABP = 180°
∴ m∠APB + 30 + 60 = 180
- Add like terms in the left hand side
∴ m∠APB + 90 = 180
- Subtract 90 from both sides
∴ m∠APB = 90°
(ii)
In Δ ADP:
∵ m∠D = 120° ⇒ Proved
∵ m∠DAP = 30° ⇒ proved
- Sum of the measures of the interior angles in a Δ is 180°
∴ m∠APD + m∠DAP + m∠D = 180°
∴ m∠APD + 30 + 120 = 180
- Add like terms in the left hand side
∴ m∠APD + 150 = 180
- Subtract 150 from both sides
∴ m∠APD = 30°
∵ m∠DAP = m∠APD = 30°
- If two angles in a triangle are equal in measures, then the triangle
is isosceles
∴ Δ ADP is an isosceles triangle
∴ AD = DP
In Δ BPC:
∵ m∠PBC = 60° ⇒ proved
∵ m∠C = 60° ⇒ proved
- Sum of the measures of the interior angles in a Δ is 180°
∴ m∠BPC + m∠PBC + m∠C = 180°
∴ m∠BPC + 60 + 60 = 180
- Add like terms in the left hand side
∴ m∠BPC + 120 = 180
- Subtract 120 from both sides
∴ m∠BPC = 60°
∵ m∠PBC = m∠C = m∠BPC = 60°
- If the three angles of a triangle are equal in measure, then
the triangle is equilateral
∴ Δ BPC is an equilateral triangle
∴ PB = PC = BC
(iii)
∵ AD = BC ⇒ opposite sides in parallelogram
∵ AD = DP ⇒ Proved
- Equate the two right hand sides of AD
∴ BC = DP
∵ BC = PC
- Equate the right hand sides of BC
∴ DP = PC
∵ DC = DP + PC
∵ DP = AD
∴ PC = AD
- Substitute DP by AD and PC by AD in CD
∴ CD = AD + AD
∴ DC = 2 AD
Learn more:
You can learn more about parallelogram in brainly.com/question/6779145
#LearnwithBrainly
We have that
7x^2 + 4x - 8 = 14-----------------> 7x²+4x-22=0
Step 1
Solving<span> a </span><span>second degree equation
</span>x1=[-b+√(b²-4ac)]/2a
x2=[-b-√(b²-4ac)]/2a
a=7
b=4
c=-22
then
x1=[-4+√(4²-4*7*(-22))]/(2*7)-----------> [-4+√(632)]/(14)
x1=-(-4+25.14)/14=1.51
x2=[-4-√(4²-4*7*(-22))]/(2*7)-----------> [-4-√(632)]/(14)
x2=-(-4-25.14)/14=-2.08
the solutions of the system is
x1=1.51
x2=-2.08
<span>to check the result
</span>
using a graph tool
see the attached figure
the answer is
the numerical expressions is
x1=[-b+√(b²-4ac)]/2a
x2=[-b-√(b²-4ac)]/2a
7x^2 + 4x - 8 = 14-----------------> 7x²+4x-22=0
Step 1
Solving<span> a </span><span>second degree equation
</span>x1=[-b+√(b²-4ac)]/2a
x2=[-b-√(b²-4ac)]/2a
a=7
b=4
c=-22
then
x1=[-4+√(4²-4*7*(-22))]/(2*7)-----------> [-4+√(632)]/(14)
x1=-(-4+25.14)/14=1.51
x2=[-4-√(4²-4*7*(-22))]/(2*7)-----------> [-4-√(632)]/(14)
x2=-(-4-25.14)/14=-2.08
the solutions of the system is
x1=1.51
x2=-2.08
<span>to check the result
</span>
using a graph tool
see the attached figure
the answer is
the numerical expressions is
x1=[-b+√(b²-4ac)]/2a
x2=[-b-√(b²-4ac)]/2a