Q ABC is a right triangle at B , BP perpendicular to AC , PR perpendicular to BC and PQ - Maths - Triangles

Question

Q ABC is a right triangle at B , BP perpendicular to AC , PR
perpendicular to BC and PQ perpendicular to AB are drawn Prove that
(i) AQ X PR = PQ2 (ii) PQ X RC = RP2 ?

Ashwini Kumar · Accepted Answer

i In △AQP and △PQB,∠PAQ=∠BPQ           ∴∠PAQ=90-∠APQ=∠BPQ∠AQP=∠PQB   both equal to 90°Therefore, by A-A similarity △AQP~△PQB⇒AQPQ=PQBQ⇒AQ×BQ=PQ2Now, BQPR is a rectangle,∴BQ=PRHence, AQ×PR=PQ2ii In △PQB and △PRC,∠BPQ=∠CPB           ∴∠BPQ=90-∠BPR=∠CPR∠PQB=∠PRC   both equal to 90°Therefore, by A-A similarity △PQB~△PRC⇒PQPR=QBRC⇒PQ×RC=PR×QBNow, BQPR is a rectangle,∴BQ=PRHence, PQ×RC=PR2

Q.ABC is a right triangle at B , BP perpendicular to AC , PR perpendicular to BC and PQ perpendicular to AB are drawn . Prove that (i) AQ X PR = PQ2 (ii) PQ X RC = RP2 ?