Question

In the given figure, O is a point in the interior of square ABCD such that ΔOAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle.

Answer 1

Given: In square ABCD, ΔOAB is an equilateral triangle.

To prove: ΔOCD is an isosceles triangle.

Proof:

$$\begin{gathered} \because \angle DAB=\angle CBA=90°\left(\mathrm{Angles}\mathrm{of}\mathrm{square}ABCD\right) \\ \mathrm{And},\angle OAB=OBA=60°\left(\mathrm{Angles}\mathrm{of}\mathrm{equilateral}∆OAB\right) \\ \therefore \angle DAB-\angle OAB=\angle CBA-\angle OBA=90°-60° \\ \Rightarrow \angle OAD=\angle OBC=30°.....\left(\mathrm{i}\right) \end{gathered}$$

Now, in ΔDAO and ΔCBO,

AD = BC (Sides of square ABCD)
$\angle$DAO = $\angle$CBO [From (i)]
AO = BO (Sides of equilateral ΔOAB)

$\therefore$ By SAS congruence criteria,
ΔDAO $\cong$ ΔCBO

So, OD = OC (CPCT)
Hence, ΔOCD is an isosceles triangle.