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$O$ is the point of intersection of two equal chords $AB$ and $CD$ such that $OB=OD$, then prove that $\vartriangle OAC$ and $\vartriangle ODB$ are similar."

Given: In the given fig. $O$ is the point of intersection of two equal chords $AB$ and $CD$ such that $OB = OD$.

To do: To prove that $\vartriangle OAC$ and $\vartriangle ODB$.

Solution: 


It is known that when two chords inside the circle intersect each other then the product of their segments is always equal.

$\Rightarrow  AO.OB=CO.OD$

$\Rightarrow  AO=CO$    ( \because OB=OD)$

In $\vartriangle AOC$,

$\angle ACO+\angle AOC+\angle CAO=180^o$

$\Rightarrow 2\angle ACO+45^o=180^o \Rightarrow  \angle ACO=67.5^o$       $(\because\ AO=OC)$

In $\vartriangle BOD$,

$\angle BOD=\angle COA=45^o$    $( vertically\ opposite\ angles)$

$\angle BDO+\angle BOD+\angle DOB=180^o$

$\Rightarrow 2\angle BDO+45^o=180^o \Rightarrow  \angle BDO=67.5^o$

In $\vartriangle AOC$ and $\vartriangle BOD$

  

$\angle ACO=\angle BDO$      $( proved\ above)$

$\angle BOD=\angle COA$      $( vertically\ opposite)$

Hence, by $AA$ similarity, 

$\vartriangle AOC\sim \vartriangle BOD$

 

Updated on: 10-Oct-2022

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