Question

In ΔACD, side BE || side CD, find BC​

Answer 1

Answer:

In ΔACD & ΔBCA,

∠BAC=∠ADC (Given)

∠ACD=∠BCA (Same angle)

=> ΔACD ≈ ΔBCA (AA)

Since, both triangles are similar ,since two corresponding angles are equal .

So, we used similarity criterion AA (angle angle) , to enhance similarity of triangle ACD and BCA .

So,Coresponding sides of similar triangles are proportional .

\begin{gathered} \frac{AC}{BC} = \frac{CD}{C A} \\ \\ =\ \textgreater \ AC^{2} = BC *CD \end{gathered}

BC

AC

=

CA

CD

= \textgreater AC

2

=BC∗CD

Hence Proved , \begin{gathered}AC^2=BC*CD \\ CA^2=CB*CD\end{gathered}

AC

2

=BC∗CD

CA

2

=CB∗CD

Answer 2

Step-by-step explanation:

In ΔACD & ΔBCA,

∠BAC=∠ADC (Given)

∠ACD=∠BCA (Same angle)

=> ΔACD ≈ ΔBCA (AA)

Since, both triangles are similar ,since two corresponding angles are equal .

So, we used similarity criterion AA (angle angle) , to enhance similarity of triangle ACD and BCA .

So,Coresponding sides of similar triangles are proportional .

\begin{gathered} \frac{AC}{BC} = \frac{CD}{C A} \\ \\ =\ \textgreater \ AC^{2} = BC *CD \end{gathered}

BC

AC

=

CA

CD

= \textgreater AC

2

=BC∗CD

Hence Proved , \begin{gathered}AC^2=BC*CD \\ CA^2=CB*CD\end{gathered}

AC

2

=BC∗CD

CA

2

=CB∗CD