Show that the ASCO is rectangular

Show that the ASCO is rectangular.

#Geometry

Note by Aly Ahmed
5 years ago

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Comments

What is "mesoparallili"?

Pi Han Goh - 4 years, 12 months ago

Since ∠OPS=90∘\angle OPS = 90^\circ and ∠OAS=90∘ \angle OAS = 90^\circ , hence ASPO ASPO is a cyclic quad.

Since ∠PAB=90∘−∠PBA=∠PQO=180∘−∠OQP \angle PAB = 90^\circ - \angle PBA = \angle PQO = 180^\circ - \angle OQP , hence APQO APQO is a cyclic quad.

Hence, ASPQOASPQO is a cyclic quad. Thus, since ∠SAO=90∘ \angle SAO = 90^\circ , hence ∠SQO=90∘ \angle SQO = 90 ^ \circ . And since ∠AOQ=90∘ \angle AOQ = 90^\circ hence ∠ASQ=90∘ \angle ASQ = 90^ \circ . So we have ASQOASQO is a rectangle.

Calvin Lin Staff - 4 years, 11 months ago

With O as the origin, let's assume the circle to be: x²+ y² = R² and point P to be (a,b). Likewise, let Q be (q,0) & S be (p,R). Now it will be adequate to show that p=q for ASQO to be a rectangle. The tangent at P is:ax+by=R² and this meets the tangent at A with p as given by: ap +bR=R² or p =R(R-b)/a. By similarity, we further claim that: R/(R+b) = q/a or q=aR/(R+b). Now p=q if R(R-b)/a = aR/(R+b) or if (R--b)(R+b) = a² or if R² - b² = a² or if a² + b² = R² which is true since P:(a,b) lies on the circle; hence p = q. In other words, ASQO is rectangular.

Ajit Athle - 4 years, 11 months ago
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