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

Question

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

Uniteasy Admin · Accepted Answer

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

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