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Extend EA to T, CA to P, (P, T on circle O)Draw OG ꓕ AP => OP = AC/2
https://photos.app.goo.gl/z8XEgvg4x6Eq1rML8see sketch for position of points M and Nnote that OAQ is isosceles right triangle Let M and N are the projection of Q and O over ACTriangle ONA congruent to AMQ ( case ASA)So ON= AM= ½. AEArea S= ½. ON.AC= ½ x ½ x AC. AC= ¼ S1
Extend AQ to meet Circle Q at X. Drop a perpendicular from C to OA meeting it at Y.Let OA = AQ= QX = r and let AC = yTr.s ACY and ACX are similar since < XAY = < ACX = 90So y/2r = h/y and h = y^2/2r = S1/2r i.e. S1 = 2rh …(1)But S = rh/2...(2)From (1) and (2) S1 = 4SSumith PeirisMoratuwaSri Lanka
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Extend EA to T, CA to P, (P, T on circle O)
ReplyDeleteDraw OG ꓕ AP => OP = AC/2
https://photos.app.goo.gl/z8XEgvg4x6Eq1rML8
ReplyDeletesee sketch for position of points M and N
note that OAQ is isosceles right triangle
Let M and N are the projection of Q and O over AC
Triangle ONA congruent to AMQ ( case ASA)
So ON= AM= ½. AE
Area S= ½. ON.AC= ½ x ½ x AC. AC= ¼ S1
Extend AQ to meet Circle Q at X. Drop a perpendicular from C to OA meeting it at Y.
ReplyDeleteLet OA = AQ= QX = r and let AC = y
Tr.s ACY and ACX are similar since < XAY = < ACX = 90
So y/2r = h/y and h = y^2/2r = S1/2r i.e. S1 = 2rh …(1)
But S = rh/2...(2)
From (1) and (2) S1 = 4S
Sumith Peiris
Moratuwa
Sri Lanka