In $$ riangle ABC$$ , $$D, E$$ are two points on the sides of $$AC$$ and $$AB$$. $$ BD$$ and $$CE$$ intersects at point $$O$$ , $$\angle OBC=\angle OCB=\dfrac{1}{2}\angle A.$$ verify that $$BE=CD. $$

Question

In  $$	riangle ABC$$ , $$D, E$$ are two points on the sides of $$AC$$ and $$AB$$. $$ BD$$ and $$CE$$ intersects at point $$O$$ , $$\angle OBC=\angle OCB=\dfrac{1}{2}\angle A.$$ verify that  $$BE=CD.     $$

Uniteasy Admin · Accepted Answer

Locate point $$F$$ on the line of $$BD$$ such that $$OF=OE$$,connect $$FC$$.
$$\because  \angle OBC=\angle OCB$$
$$	herefore  OB=OC $$
$$在	riangle EOB与	riangle FOC中 $$
```katex
\begin{cases} 
EO =FO \ 
\angle EOB=\angle FOC  \
OB=OC
\end{cases}
```
$$	herefore  	riangle EOB≌	riangle FOC $$
$$	herefore BE=CF$$n
$$\angle EBO=\angle FCO $$
$$\because \angle FDC=\angle A+\angle EBO $$n
$$\angle DFC=\angle OBC+\angle FCB  $$
$$\angle  DFC=\angle OBC+\angle OCB+\angle FCO$$
$$\angle  DFC=2\angle OBC+ \angle FCO$$n
$$\because  \angle OBC=\angle OCB=\dfrac{1}{2}\angle A$$
$$ 	herefore   \angle A=2 \angle OBC=2\angle OCB $$
Substitute to(3), and since $$\angle EBO=\angle FCO $$,
Therefore,$$  \angle  DFC=\angle A+\angle EBO$$
$$	herefore   \angle  DFC=\angle FDC $$
$$	herefore  CF=CD $$
$$\because  BE=CF$$
$$	herefore  BE=CD $$

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