Mathematical Enigma- V

Geometry Level 3

Let ABC be an isosceles triangle ( $AB = AC)$ with $\angle BAC = 20°$ . Point $D$ is on side $AC$ such that $\angle DBC = 60°$ . Point $E$ is on side $AB$ such that $\angle ECB=50°$ . Find the measure of $\angle EDB$ .

The answer is 30.

1 solution

Carlos Victor
Oct 19, 2015

Observe that angle(BEC)=50°, angle(BDC)=40° and BE=BC. Take the point "F" on DC, such that angle(CBF)=20°. Observe that the triangle BFC is isosceles with BC=BF. Then BF=BE with angle(EBF)=60°, and the triangle BEF is equilateral. As the angle(BDC)=40°, then the triangle BFD is isosceles, so BF=FD, and since EF=BF=BE then DF=EF. The angle(EFD)=40° since angle(BFC)=80°, and the triangle EFD is isosceles; so angle(EDF)=70° and we will angle(EDB)= 30°.

Mathematical Enigma- V

The answer is 30.

1 solution

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