Question

ABCD is a square piece of paper. M and N are the respective midpoints of AB and CD. P is a point on AM such that if the piece of paper is folded along DP, then A lands on a point Q on the segment MN. What is the degree of `angle`ADP ?

Answer 1

`angle ADP=15^@`

Explanation:

As shown in the figure `ABCD` is a square with side length `=x`,
given `M and N` are the midpoints of `AB and CD`, respectively,
`=> DN=1/2x`
if the paper is folded along `DP`, `A` lands on `Q` on segment `MN`,
`=> AP=PQ`
`=> DQ=AD=x`
`DeltaDAP and DeltaDQP` are congruent
and `cos alpha=(DN)/(DQ)=(1/2x)/x=1/2`
`=> alpha=cos^-1(1/2)=60^@`
`=> angleADQ=2beta=90-60=30^@`
`=> angleADP=beta=1/2*30=15^@`