From triangle QRS let
a = /RSQ, b = /QRS, c = /SQR
From triangle PQS let
d = /PQS, e = /SPQ, f = /QSP
a + b + c = 180 = d + e + f
= b + (a + f) + f
==> c = 2*f
Let T be the point on side RS such that QT bisects /SQR.
QR*RS
Therefore, RT = --------- and
QR + QS
QT^2 = QR*QS*sqrt[1 - (RS/[QR + QS])^2] (1)
Checking the angles we see that triangles RTQ and QPS
are similar. Therefore,
RT QT QR
---- = ---- = ---- (2)
QP SP QS
Combining (1) and (2) gives
RS = QP*(QR + QS)/QS and
SP = sqrt[ QS*(QS^2 - QP^2)/QR ]
QED
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Posted by Bractals
on 2015-11-01 09:43:08 |
Solution