CBSE Class 10 Answered
pls
Asked by garimasingh | 13 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 | 14 Oct, 2009, 10:16: AM
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