Question
In the figure, BD bisects ∠ABC , point D resides on BD such that ∠A+∠C=180° . Prove that AD=DC
Answer
Drop altitudes from D to the two legs of the angle \angle ABC
Since BD bisects \angle ABC ,
DE = DF
(1)
In the right triangle \triangle ADE
\angle ADE+ \angle DAE = 90 \degree
\angle ADE+ (180\degree -\angle A ) = 90 \degree
\angle A -\angle ADE =90\degree
(2)
In the right triangle \triangle DCF
\angle C+\angle FDC = 90\degree
(3)
Addition of (2) and (3) gives
\angle A+\angle C+\angle FDC -\angle ADE=180\degree
Substituting the given condition ∠A+∠C=180° yields
\angle FDC =\angle ADE
(4)
According to (1) and (4), \triangle ADE and \triangle DCF are congruent triangles.
\triangle ADE \cong \triangle DCF AD and DC are their corresponding sides. Hence,
AD = DC