Question

seg $AB$ and seg $CD$ intersect each other at point $M$.If $AM\times MD$ $=CM\times MB$ then show that $\angle CAB\cong \angle DBA$

Answer 1

Given: AM\times MD=CD \times MB$

$\frac{AM}{CM}=\frac{MB}{MD}$

To prove: $\angle CAB=\angle DBA$

In $△AMC$ and $△BMD$

$\frac{AM}{MC}=\frac{BM}{MD}$ & $\angle AMC=\angle BMD$

$△AMC\sim △MBD$ [by $SAS$ similarity ]

By corresponding parts of similar triangle;s.

$\angle CAM=\angle DBM$

$\Rightarrow \angle CAB=\angle DBA$

Hence prove