If AB=AC of ABC Isosceles triangle and P is any one point on the extended BC arm. PQ and PR are perpendicular to the sides of point AB and AC from point p, respectively. Perpendicular BS from point B to AC prove PQ - PR = BS.
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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
Hope this helps you ☺️☺️⭐✨✨⭐✌️✌️
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