1) ABC is an isosceles triangle with AB=AC. P is any point on BC. perpendiculars PQ and PR are drawn to sides AB and AC respectively. also CS perpendicular to AB. prove that CS=PQ+PR.

2) M and N are midpoints of sides DC and AB respectively of parallelegram ABCD.show that ar(BEC) =ar(MCBN) when it is given that BD parallel to EC.

3) In triangle ABC P and Q are the midpoints of sides AB and AC respectively.R and S are the midpoints of PC and PB respectively.prove that BQ and SR bisect each other.

Given    ABC is an isosceles triangle with AB=AC .PQAB,PRAC,CSAB
To prove  CS =PQ+PR

As AB=AC  hence ABC =ACB(angles opposite to equal sides of a triangle are equal)Now in PQB  and PRC,we havePBQ=PCR PQB=PRC (=90°)QPB=RPC(remaining angle)Hence by AAA similarity ,PQB  ~ PRCPQPR=PBPCPQPR+1=PBPC+1PQ+PRPR=PB+PCPCPQ+PR =BCPC×PR ...........(1)Now in  CSB  and PRC,we haveCBS=PCR CSB=PRC (=90°)SCB=RPC(remaining angle)Hence by AAA similarity ,CSB   PRCCSPR=BCPCCS =BCPC×PR ........(2)From (1) and (2) we can say that  PQ+PR = CS  proved 

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