Q90 of 93 Page 6

ABCDE is a regular pentagon and bisector of ∠ BAE meets CD at M. If bisector of ∠ BCD meets AM at P, find ∠ CPM.


Given: ABCDE is a regular pentagon. The bisector of ∠ BAE meets CD at M and bisector of ∠ BCD meets AM at P.
To find: The measure of ∠ CPM.
Proof: Each angle of a pentagon =
=
=
=
=
= 108°
In quadrilateral ABCM,
∠ BAM + ∠ B + ∠ C + ∠ CMP = (2n - 4) right angles
= (2 × 4 - 4) right angles
= 4 right angles
= 4  × 90°
= 360°
⇒ +108° +108° +∠ CMP = 360°
                 ⇒ 270° +∠ CMP = 360°
                          ⇒ ∠ CMP = 360° -270°
                          ⇒ ∠ CMP = 90°
∠ PCM = ∠ BCM = × 108° = 54°
In Δ CPM, ∠ CPM + ∠ PCM + ∠ CMP = 180° [Sum of the three angles of a triangle is 180° ]
          ⇒ ∠ CPM + 54° + 90° = 180°
                 ⇒ ∠ CPM + 144° = 180°
                            Þ ∠ CPM = 180° - 144° = 36°

More from this chapter

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88 Prove that the sum of the angles of a hexagon is 720° . 89 In a ΔABC, ∠ B = 45°, ∠C = 55° and bisector of ∠ A meets BC at a point D. Find ∠ ADB and ∠ ADC. 91 In figure, bisectors of ∠ B and ∠ D of a quadrilateral ABCD meet CD and AB produced at P and Q respectively. Prove that ∠ P +∠ Q=(∠ ABC + ∠ ADC).
           92 Which of the following statements are true (T) and which are false (F)? (i)An exterior angle of a triangle is less than either of its interior opposite angles. (ii) Sum of the three angles of triangle is 180° . (iii) Sum of the four angles of a quadrilateral is three right angles. (iv) A triangle can have to right angles. (v) A triangle can have two acute angles. (vi) A triangle can have two obtuse angles. (vii) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.