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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
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Solution
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°.
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