) Draw a rectangle ABCD with base 4 cm and diagonal BD = 5 cm. Measure CB. Take any
point P on AD. Join BP. From C draw a line
parallel to BP to meet AP produced at Q.
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According to question
in triangle DQC and triangle PAB
CD = AB (sides of rectangle)
CQ is parallel to BP
Also
Angle QCD = Angle PBC
thus
traingles are similar
so we can say that
PBQC is a parallelogram
in triangle DQC and triangle PAB
CD = AB (sides of rectangle)
CQ is parallel to BP
Also
Angle QCD = Angle PBC
thus
traingles are similar
so we can say that
PBQC is a parallelogram
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