Short Answer Type In a quadrilateral ABCD, the line segments bisecting ∠C and ∠D meet at E.
Prove that ∠A + ∠B = 2 ∠CED
Given: In a quadrilateral ABCD, the line segments bisecting ∠C and ∠D meet at E.]
To Prove: ∠A + ∠B = 2∠CED
Proof: In quadrilateral ABCD,
∠A + ∠B + ∠C + ∠D = 360° ...(1)
| Angle sum property of a quadrilateral
In ∆CED,
∠CED + ∠EDC + ∠ECD = 180°
| Angle sum property of a triangle
⇒ 2 ∠CED + ∠D + ∠C = 360° ...(2)
From (1) and (2),
2 ∠CED + ∠D + ∠C = ∠A + ∠B + ∠C + ∠D
⇒ 2 ∠CED = ∠A + ∠B
⇒ ∠A + ∠B = 2 ∠CED
ABC is an isosceles triangle in which AB = AC. AD bisects ∠PAC and CD || AB. Show that
(i) ∠DAC = ∠BCA
(ii) ABCD is a parallelogram
ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle ∠PAC and CD || AB (see figure). Show that:
(i) ∠DAC = ∠BCA and
(ii) ABCD is a parallelogram.
ABCD is a parallelogram in which P and Q are mid-points of opposite sides AB and CD (see figure). If AQ intersects DP at S and BQ intersects CP at R, show that:
(i) APCQ is a parallelogram.
(ii) DPBQ is a parallelogram.
(iii) PSQR is a parallelogram.
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