Question

4. In the adjoining figure, ABCD is a
parallelogram in which LA = 60°. If the
bisectors of ZA and B meet DC at P,
prove that (i) ZAPB = 90°, (ii) AD = DP and
PB = PC = BC, (iii) DC = 2AD.
​

Answer 1

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 = $\frac{1}{2}$ m∠A

∴ m∠BAP = m∠DAP = $\frac{1}{2}$ (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 = $\frac{1}{2}$ m∠B

∴ m∠ABP = m∠CBP = $\frac{1}{2}$ (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

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