Question

4.
In an isosceles triangle ABC, AB=AC and P is any point on produced side BC. PQ and
PR are perpendicular on sides AB and AC from the point Prespectively. BS is perpendicular
on side AC from point B; let's prove that PQ-PR = BS.
​

Answer 1

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

Hope this helps you ☺️☺️⭐✨✨⭐✌️✌️