Solve the problems. Write the complete proof.
ONLY HAVE TO PROVE HOW BD IS PERPENDICULAR TO AC!

A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. An isosceles triangle is one with two equal-length sides.
A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. The sum of all the angle of a triangle is always equal to 180°.
In ΔBAD and ΔBDC,
m∠ADB = m∠CDB {Given}
BD = BD {Common side in two triangles}
AD ≅ DC {Given}
Using the ASA postulate the two triangles are congruent. Therefore,
m∠BAD= m∠BCD.........equation 1
m∠ABD= m∠CBD
In ΔADC, AD ≅ DC {Given}, therefore, the triangle is an isosceles triangle. Thus,
m∠DAE = m∠DCE.......equation 2
Adding two of the equation 1 and 2,
m∠BAD + m∠DAE = m∠DCE + m∠BCD
∠BAC = ∠BCA
Since in ΔABC, ∠BAC = ∠BCA therefore, the triangle is an isosceles triangle, thus, AB = BC,
BE is the common side between ΔAEB and ΔCEB
Also, m∠ABD= m∠CBD
Therefore, ΔAEB ≅ ΔCEB
Now as ΔAEB ≅ ΔCEB, therefore, AE=EC,
Thus, BE is the median of the isosceles triangle ABC, and for an isosceles triangle, the median opposite to the non-common sides is the perpendicular bisector to the opposite side.
Hence, BD⊥AC.
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