Solution of Triangle using Circumcircle

In triangle ABC , let D be the mid point of BC. If ∠ADB = 45° and ∠ACD = 30° , determine ∠BAD.
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Draw BL perpendicular to AC and join L to D. Since ∠BCL = 30° , we get ∠CBL = 60°.
Since BLC is a right triangle with ∠BCL = 30° , we have BL = BC /2 = BD. Thus in triangle BLD , we observe that BL = BD and ∠DBL= 60°.

This implies that BLD is an equilateral triangle hence LB = LD. Using ∠LDB = 60° and ∠ADB=45° , we get ∠ADL = 15°. But ∠DAL = 15° , thus LD = LA. Hence we have LD = LA = LB.

This implies that L is the circumcenter of the triangle BDA , thus ∠BAD = ∠BLD / 2 = 60° / 2 = 30°

Circumcircles through a common point

Let ABC be a triangle and let P, Q, R be any points on the sides BC, CA, AB, respectively. Then the circumcircles of ARQ, BP R, CQP pass through a common point. This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Medial Triangle - III

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Nine Point Circle

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Circumcentre of a Triangle

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