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Q14  Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and
        median PM of another triangle PQR. Show that \Delta ABC \sim \Delta PQR

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\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}              (given)

Produce AD and PM to E and L such that AD=DE  and PM=DE. Now,

join B to E,C to E,Q to L and R to L.

AD and PM are medians of a triangle, therefore

QM=MR  and    BD=DC

AD = DE            (By construction)

PM=ML               (By construction)

So, diagonals of ABEC bisecting each other at D,so ABEC is a parallelogram.

Similarly, PQLR is also a parallelogram.

Therefore, AC=BE ,AB=EC and PR=QL,PQ=LR

\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}               (Given )

\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{2.AD}{2.PM}

\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{AE}{PL}

  \Delta ABE \sim \Delta PQL                (SSS similarity)

\angle BAE=\angle QPL ...................1           (Corresponding angles of similar triangles)

Similarity, \triangle AEC=\triangle PLR

\angle CAE=\angle RPL........................2

Adding  equation 1 and 2,

\angle BAE+\angle CAE=\angle QPL+\angle RPL

\angle CAB=\angle RPQ............................3

In \triangle ABC\, and\, \, \triangle PQR,

\frac{AB}{PQ}=\frac{AC}{PR}        ( Given )

\angle CAB=\angle RPQ        ( From above equation 3)

\triangle ABC\sim \triangle PQR         ( SAS similarity)     

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Posted by

seema garhwal

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