In the adjoining figure, ABCD is a parallelogram, E is midpoint of AB and CE bisects `angle BCD`. Prove that

(i) AE = AD (ii) DE bisects `angle ADC ` and (iii) `angle DEC = 90^(@)`.

Explanation

AB||DC and EC cuts them `rArr angle BEC = angle ECD`
`rArr angle BEC = angle ECB " " [ because angle ECD = angle ECD`
`rArr EB = BC rArr AE = AD.`
Now, `AE= BC rArr angle ADE = angle AED rArr angle ADE = angle EDC " " [because angle AED=angle EDC ( " alt. interior " angle )].`
`therefore " DE bisects " angle ADC.`
Further , `angle ADC + angle BCD = 180^(@) " " [ "co-interior " angle ]`
`rArr (1)/(2) angle ADC +(1)/(2) angle BCD = 90^(@) rArr angle EDC + angle DCE = 90^(@).`
But, ` angle EDC + angle DCE + angle DEC = 180^(@) " " [ "sum of the " angle " of a " triangle ]`
`therefore angle DEC =90^(@).`