Question

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 ab. Prove that cs=pq+pr

Answer 1

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 ☺️☺️⭐✨✨⭐✌️✌️

Answer 2

Answer:

Step-by-step explanation:

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