Geometry; Triangle Congruences

[MadisonNeedsHelp]

Please tell me what step fives reason is, thanks <3

 

[MadisonNeedsHelp]

Oh my, I just realized, thank you. It looks like this; your the bomb.

 

[soroban]

Hello, MadisonNeedsHelp!

Please tell me what step five's reason is. Thanks.

            C           E 
            *           *
           * *         * * 
          *   *       *   *
         *     *     *     *
        *       *   *       *
       *         * *         *
    A *  *  *  *  *  *  *  *  * D
                  B

\(\displaystyle \text{Given: }\:B\text{ is the midpoint of }AD.\quad \angle ACB = \angle BED. \quad BC \parallel AD\)

\(\displaystyle \text{Prove: }\Delta BAC \,\cong\,\Delta DBE\)


\(\displaystyle \begin{array}{cccccccc}1. & B\text{ midpt. of }AD && 1. & \text{Given} \\ 2. & AB \,=\,BD && 2. & \text{by Midpoint} \\ 3. & \angle ACB \,=\,\angle BED && 3. & \text{Given} \\ 5. & BC \,=\,DE && 5. & ? \\ 6. & \Delta BAC \,\cong\,\Delta DBE && 6. & \text{by SAS} \end{array}\)

I see no justification for statement 5.
Perhaps there is a typo in the problem.


We can provide a proof, though.

\(\displaystyle \text{In }\Delta ABC\text{ and }\Delta BDE\!:\;\;\begin{Bmatrix}\angle ACB \:=\:\angle BED && \text{Given} \\ \angle CBA \:=\:\angle EDB && \text{corres. angles} \end{Bmatrix}\)

\(\displaystyle \text{Hence, their third angles are equal: }\:\angle CAB \,=\,\angle EBD\)


\(\displaystyle \text{Therefore: }\: \Delta BAC \,\cong\,\Delta DBE\,\text{ by ASA.}\)

 

[aesha12]

Geometry Triangles congruence


Given: <1=<2 ; Segment TP= Segment RA
Prove: <3=<4
If u will click on the pic the pic will enlarge.

Anybody plzz help me !