Post History
#2: Post edited
by
DonielF
·
2020-10-05T02:25:00Z (over 3 years ago)
- For a triangle defined by three points $X, Y, Z$:
- $$A_{\triangle XYZ}=\frac12\cdot\overline{XY}\cdot\overline{YZ}\cdot\sin{\angle Y}$$
- Since $A_{\triangle ACD}=A_{\triangle AEB}$,
- $$\overline{AB}\cdot\overline{BE}\cdot\sin\angle B=\overline{AC}\cdot\overline{CD}\cdot\sin\angle C$$
- Since $\overline{AB}\cong\overline{AC}$ and $\angle B\cong\angle C$, this equation simplifies to
- $$\overline{BE}=\overline{CD}$$
From there congruence of the triangles follows from SAS, and congruent parts of congruent triangles are congruent. $\square$
- For a triangle defined by three points $X, Y, Z$:
- $$A_{\triangle XYZ}=\frac12\cdot\overline{XY}\cdot\overline{YZ}\cdot\sin{\angle Y}$$
- Since $A_{\triangle ACD}=A_{\triangle AEB}$,
- $$\overline{AB}\cdot\overline{BE}\cdot\sin\angle B=\overline{AC}\cdot\overline{CD}\cdot\sin\angle C$$
- Since $\overline{AB}\cong\overline{AC}$ and $\angle B\cong\angle C$, this equation simplifies to
- $$\overline{BE}=\overline{CD}$$
- From there congruence of the triangles follows from SAS, and the rest follows as you set out in your original post. $\square$
#1: Initial revision
by
DonielF
·
2020-10-05T02:23:59Z (over 3 years ago)
For a triangle defined by three points $X, Y, Z$:
$$A_{\triangle XYZ}=\frac12\cdot\overline{XY}\cdot\overline{YZ}\cdot\sin{\angle Y}$$
Since $A_{\triangle ACD}=A_{\triangle AEB}$,
$$\overline{AB}\cdot\overline{BE}\cdot\sin\angle B=\overline{AC}\cdot\overline{CD}\cdot\sin\angle C$$
Since $\overline{AB}\cong\overline{AC}$ and $\angle B\cong\angle C$, this equation simplifies to
$$\overline{BE}=\overline{CD}$$
From there congruence of the triangles follows from SAS, and congruent parts of congruent triangles are congruent. $\square$