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As shown in the figure, in △ ABC, ad is the middle line on BC, e is the point on ad, and CD = CE, ∠ EAC = ∠ B, try to explain △ AEC ∽ BDA
Because CD = CE
So angle CED = angle CDE
Because CED + CEA = 180, CDE + BDE = 180
So CEA = ADB
From the sum of the internal angles of the triangle equal to 180 degrees, DAB = ECA
Because ∠ EAC = ∠ B
So △ AEC ∽ BDA
As shown in the figure, in the triangle ABC, the angle c is 90 degrees, D is the midpoint of the hypotenuse AB, De is perpendicular to AB and BC is perpendicular to e. the known angle EAC is 2 to 5 than the angle DAE
Finding the degree of BAC
D is the midpoint of the hypotenuse AB, De is perpendicular to AB and BC is perpendicular to E,
△BDE≌△ADE,
∠DBE=∠DAE.
Let ∠ EAC = 2x, then ∠ DAE = 5x, ∠ DBE = 5x,
Angle c is 90 degrees, ∠ cab + ∠ CBA = 90 degrees,
2x+5x+5x=90,
x=7.5,
∠BAC=7x=52.5°.
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