Question

P is a point on the AD mean of ∆ABC. Increased BP and CP intersect AC and AB at point Q and R, respectively. To prove That, RQ || BC.​

Answer 1

Answer:

Given: ABC is an equilateral triangle. P, Q, R are points on AB, BC, CA respectively, such that

AP=BQ=CR

We know, AB=BC=CA

AP=BQ=CR

AB−AP=BC−BQ=CA−CR

BP=CQ=AR (I)

Now, In △APR and △PBQ

∠A=∠B=60  

∘

 

AP=BQ (Given)

BP=AR (From I)

Thus, △APR≅△BQP (SAS rule)

Hence, PQ=PR (By cpct)...(II)

Similarly, △APR≅△CRQ

hence, PR=QR ..(III)

thus, from II and III

PQ=QR=PR

∴△PQR is an equilateral triangle.

Step-by-step explanation: