Question

In quad^ ABCD sideBC is parallel to sideAD.diagonal AC and diagonal BD intersect in point Q,if AO=1/3 AC then show that DQ=1/2BQ...plz guys help me.....​

Answer 1

Answer:

AQ= AC/3

AQ/QC = 1/2

AQ+QC = AC

AC/3 + QC = AC

QC = AC - AC/3

QC = 2/3 AC

In triangle QAD and triangle QCB

angle QAD =angle QCB (opposite alternate angle)

angle QDA = angle QBC

angle AQD = angle CQB

∆ QAD ~ ∆QCD

AQ/QD = CQ/QB

AQ/CQ = QD/QB

$\frac{ \frac{AC}{3} } {\frac{2AC}{3} } = \frac{QD}{QB}$

QD/QB = 1/2

QD = QB/2

QD = 1/2 BQ

hence proved

Explanation:

hope it helps yrr

Answer 2

Explanation:

AQ= AC/3

AQ/QC = 1/2

AQ+QC = AC

AC/3 + QC = AC

QC = AC - AC/3

QC = 2/3 AC

In triangle QAD and triangle QCB

angle QAD =angle QCB (opposite alternate angle)

angle QDA = angle QBC

angle AQD = angle CQB

∆ QAD ~ ∆QCD

AQ/QD = CQ/QB

AQ/CQ = QD/QB

\frac{ \frac{AC}{3} } {\frac{2AC}{3} } = \frac{QD}{QB}

3

2AC

3

AC

=

QB

QD

QD/QB = 1/2

QD = QB/2

QD = 1/2 BQ

hence proved