D and E are mid-points of sides AB and AC respectively of the ΔABC. A line drawn from A meets BC at H and DE at K.

AK : KH = ??

D and E are mid-points of sides AB and AC respectively of the ΔABC  

$$=$$>  DE is parallel to BC

AB = 2AD and AC = 2AE

In ΔABC and ΔADE

$$\angle$$ABC = $$\angle$$ADE

$$\angle$$ACB = $$\angle$$AED

$$\angle$$A is the common angle

$$=$$>  ΔABC is similar to ΔADE

$$=$$>  $$\frac{AB}{AD}=\frac{AH}{AK}=\frac{AC}{AE}$$

$$=$$>  $$\frac{AB}{AD}=\frac{AH}{AK}$$

$$=$$>  $$\frac{2AD}{AD}=\frac{AK+KH}{AK}$$

$$=$$>   2AK = AK + KH

$$=$$>   AK = KH

$$=$$>   AK : KH = 1 : 1

Hence, the correct answer is Option D