Question

hiw to prove theles theoram​

Answer 1

$\textbf{ \large{ \underline { \: \underline{ \: Thales Theorem : \: } \: }}}$

Thales Theorem or Basic Proportionality Theorem, the theorem states that line parallel to the one side of the triangle divides the other two sides proportionally.

$\textsf{ \underline { \: \underline{ Given } \: }}$

In a triangle ABC has a line drawn parallel to BC which cuts AB and AC at points P and Q respectively.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \textsf{Refer the attachement}}$

$\textsf{ \underline { \: \underline{ To Prove : } \: }} \frac{ \textsf{AP}}{ \textsf{PB}} = \frac{ \textsf{AQ}}{ \textsf{AC}}$

$\textsf{ \underline { \: \underline{ Construction : } \: }}$ Let the point P divides AB in ratio of l : m where l and m are natural numbers. Divide AP in (l) and PB in (m) equal parts.

$\textsf{ \underline { \: \underline{ Proof : } \: }}$ $\star$ AP and AQ are cut in l equal parts

Similarly,

$\star$ PB and QC are cuts in m equal parts.

$\rightarrow \: \frac{ \text{AP}}{ \text{PB}} = \frac{ \text{l}}{ \text{m}} \\ \rightarrow \: \frac{ \text{AQ}}{ \text{QC}} = \frac{ \text{l}}{ \text{m}}$

Therefore,

$\frac{ \text{AP}}{ \text{PB}} = \frac{ \text{AQ}}{ \text{QC}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed { \boxed{ \frac{ \text{AP}}{ \text{PB}} = \frac{ \text{AQ}}{ \text{QC}} }}$