Let \(ABC\) be a triangle with \(AB = AC\). If \(D\) is the mid point of \(BC\), \(E\) the foot of perpendicular drawn from \(D\) to \(AC\) and \(F\) the mid point of \(DE\), \(AF\) and \(BE\) are joined and meet at \(M\).

What is the measure of the angle \(AMB\)?

P.S. please post a pure geometrical proof of this (without use of coordinate geometry)

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