Question

In triangle ABC ,P,Qand R are the midpoint of sideAB, AC and BC respectively Seg AS perpendicular to side Bc. Prove that: PQRS is cyclic​

Answer 1

Answer:

P and Q are the midpoints of sides AB and AC of Δ A B C . ΔABC. ∴ ∴ by the midpoit theorem. P Q ∣ ∣ B C PQ∣∣BC ( i.e. BR ) and PQ = 1 2 B C =12BC …..(1) B R = 1 2 B C BR=12BC ….( R is the midpoint of side BC ) …(2) From (1) and (2), P Q ∣ ∣ B R PQ∣∣BR and P Q = B R PQ=BR. ∴ □ ∴□ PQRB is a parallelogram ∴ ∠ B = ∠ Q ∴∠B=∠Q ...(Opposite angles of a parallelogram ) ...(3) In right angled triangle, median to the hypotenuse is half of the hypotenuse. ∴ ∴ in Δ A B S , S P = 1 2 A B ΔABS,SP=12AB. P B = 1 2 A B PB=12AB ...( P is the midpoint of side AB ) ∴ S P = P B ∴SP=PB ∴ ∠ B = ∠ P S B ∴∠B=∠PSB ...( Isosceles triangle theorem ) ...(4) From (3) and (4), ∠ Q = ∠ P S B ∠Q=∠PSB ....(5) ∠ P S B + ∠ P S R = 180 ∘ ∠PSB+∠PSR=180∘ ....( Angles in a linear pair ) ...(6) From (5) and (6) ∠ Q + ∠ P S R = 180 ∘ ∠Q+∠PSR=180∘ ∴ ∴ by the converse of cyclic quadrilateral theorem , □ □ PQRS is a cyclic quadrilateral.Read more on Sarthaks.com - https://www.sarthaks.com/1215287/delta-abc-are-midpoints-sides-and-respectively-seg-bc-and-side-prove-that-square-pqrs-cycli?show=1215616#a1215616

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