#### Question

In the figure, BD bisects ∠ABC , point D resides on BD such that ∠A+∠C=180° . Prove that AD=DC

In the figure, BD bisects ∠ABC , point D resides on BD such that ∠A+∠C=180° . Prove that AD=DC

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