CBSE Class 10 Answered
AB is a diameter of a circle. From a, two secants AP and AQ have been drawn which intersect the tangent at B to the circle at points L and M respectively.
Prove that:-
triangle APQ is similar to traingle AML and P,Q,M,L are concyclic
Asked by rushabhjain.avv | 30 Jan, 2019, 08:20: PM
Expert Answer
Figure shows a circle with diameter AB. A tangent is drawn at B. Secants AP and AQ intersect the tangenet at L and M.
By tangent secant theorem, we have LB2 = LA×LP = LA×(LA-AP) = LA2 - LA×AP ..............(1)
eqn.(1) is written as, LA × AP = LA2 - LB2 = AB2 ...............................(2)
Similarly, by tangent secant theorem, MB2 = MA×MQ = MA×(MA-AQ) = MA2 - MA×AQ ................(3)
eqn.(3) is written as, MA × AQ = MA2 - MB2 = AB2 .........................(4)
from (2) and (4), we have, LA × AP = MA × AQ or ............................(4)
In ΔAPQ and ΔAML, PAQ is common angle and sides of the common angles are in a ratio given by eqn.(4).
Hence ΔAPQ is similar to ΔAML
in similar triangles ΔAPQ and ΔAML , we have APQ = AML ...................(5)
also AQP = ALM ......................(6)
since APQ + LPQ = 180, using eqn.(5 ), LPQ+QML = 180 .....................................( 7)
similarly, since AQP + PQM = 180, using eqn.( 6), PQM + PLM = 180 .................(8)
from (7) and (8 ), we have, PQML is cyclic quadrilateral and points P, Q, M and L are concyclic
Answered by Thiyagarajan K | 01 Feb, 2019, 08:46: AM
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