Question

In a parallelogram ABCD, E is the midpoint of AB and DE bisects angle D. Prove that:CE is the bisector of angle C and angle DEC is a right angle

Answer 1

Since BC = BE
⇒ ∠BEC = ∠BCE   ...(Angles opposite to equal sides are equal)
∠BEC = ∠ECD      ...(Alternate angles)
⇒ ∠BCE = ∠ECD
⇒ CE is the bisector of ∠C  ....(proved)

∠DCE = `(1)/(2)∠"C"`   ...(Given CE bisects ∠D)

∠CDE = `(1)/(2)∠"D"`  ...(Given DE bisects ∠D)

∠DCE + ∠CDE

= `(1)/(2)(∠"C" + ∠"D")`

= `(1)/(2) xx 180°` = 90°

Thus, in ΔDCE,
∠DEC = 180° - ∠DCE + ∠CDE = 180° - 90°
⇒ ∠DEC = 90°.