Question

ABCD is a square. The diagonals AC & BD cut at E. From B a perpendicular is drawn to the bisector of /_DCEand it cuts AC atP and DC at Q. Can you prove that DQ = 2PE?

Answer 1

see explanation.

Explanation:

Let `AB=s, => AC=BD=sqrt2s, => BE=(sqrt2s)/2`,
given `CR` is the angle bisector of `angleDCE` and `BQ` is perpendicular to `CR`,
`=> angleQCR=anglePCR=22.5^@`
`=> DeltaCRQ and DeltaCRP` are congruent,
`=> color(red)(CQ=CP)`
as `BR` is perpendicular to `CR`,
`=> angleCBR=90-22.5-45=22.5^@`
`=> angleQBE=45-22.5=22.5^@=angleCBR`,
`=> BQ` is the angle bisector of `angleCBD`,
By angle bisector theorem,
in `DeltaBCE, (CP)/(PE)=(BC)/(BE)=s/((sqrt2s)/2)=2/sqrt2 --- EQ(1)`
in `DeltaBCD, (CQ)/(DQ)=(BC)/(BD)=s/(sqrt2s)=1/sqrt2--- EQ(2)`
`(EQ(1))/(EQ(2)), => (CP)/(PE)*(DQ)/(CQ)=2/sqrt2*sqrt2=2`,
since `CQ=CP, => (DQ)/(PE)=2`
Hence `DQ=2PE`