pls

Asked by garimasingh | 13th Oct, 2009, 12:41: PM

Expert Answer:

PQ ll BC

also, AD  is median,

So

BD=DC

let AD meets PQ  in E.

In triangles,APE and ABD

angle APE=angleABD...corr angles

angle A is common , so triangles are sim  by AA rule.

so

AP/AB=AE/AD=PE/BD

simly by taking trangles AQE and ACD we get,

AQ/AC=AE/AD=QE/CD

so from above eqns we get

PE/BD=QE/CD

so

PE/QE=BD/CD

  but BD/CD=1 ...  since AD is median

so

PE/QE=1

PE=QE

E  is mid pt of PQ

AD  bisects PQ.

 

Answered by  | 14th Oct, 2009, 10:16: AM

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