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?

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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?

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Answer 1 · 2017-10-24T16:25:32

see explanation.

Explanation:

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Let $A B = s, \Rightarrow A C = B D = \sqrt{2} s, \Rightarrow B E = \frac{\sqrt{2} s}{2}$ $AB=s, => AC=BD=sqrt2s, => BE=(sqrt2s)/2$ ,
given $C R$ $CR$ is the angle bisector of $∠ D C E$ $angleDCE$ and $B Q$ $BQ$ is perpendicular to $C R$ $CR$ ,
$\Rightarrow ∠ Q C R = ∠ P C R = {22.5}^{\circ}$ $=> angleQCR=anglePCR=22.5^@$
$=> DeltaCRQ and DeltaCRP$ 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 $DQ=2PE$

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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?

1 Answer

Explanation:

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