1) ABC is an isosceles triangle with AB=AC P is any point on BC perpendiculars PQ and PR - Maths - Areas of Parallelograms and Triangles

Question

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

Utsav · Accepted Answer

Given ABC is an isosceles triangle with AB=AC
PQ⊥AB,PR⊥AC,CS⊥AB
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
have∠PBQ=∠PCR ∠PQB=∠PRC
(=90°)∠QPB=∠RPC(remaining angle)Hence by AAA similarity
,∆PQB ~
∆PRC∴PQPR=PBPC⇒PQPR+1=PBPC+1⇒PQ+PRPR=PB+PCPC⇒PQ+PR
=BCPC×PR (1)Now in ∆CSB and ∆PRC,we
have∠CBS=∠PCR ∠CSB=∠PRC
(=90°)∠SCB=∠RPC(remaining angle)Hence by AAA similarity
,∆CSB ∼ ∆PRC∴CSPR=BCPC⇒CS
=BCPC×PR (2)From (1) and (2) we can say that PQ+PR = CS
proved