A point D is on the side BC of an equilateral triangle

We are given $A B C$ is an equilateral triangle with $C D=\frac{1}{4} B C$

We have to prove $A D^{2}=13 D C^{2}$

Draw $A E \perp B C$

In $\triangle A E B$ and $\triangle A E D$ we have $A B=A C$

$\angle A E B=\angle A E C=90^{\circ}$

$A E=A E$

So by right side criterion of similarity we have

Thus we have

$D C=\frac{1}{4} B C$ and $B E=E C=\frac{1}{2} B C$

Since $\angle C=60^{\circ}$, therefore

$A D^{2}=A C^{2}+D C^{2}-2 D C \times E C$

$A D^{2}=A C^{2}+\left(\frac{1}{4} B C\right)^{2}-2 \times \frac{1}{4} B C \times \frac{1}{2} B C$

$A D^{2}=A C^{2}+\frac{1}{16} B C^{2}-2 \times \frac{1}{4} B C \times \frac{1}{2} B C$

$A D^{2}=A C^{2}+\frac{1}{16} B C^{2}-B C^{2}$

We know that AB = BC = AC

$A D^{2}=B C^{2}+\frac{1}{16} B C^{2}-B C^{2}$

$A D^{2}=\frac{16 B C^{2}+1 B C^{2}-4 B C^{2}}{16}$

$A D^{2}=\frac{13}{16} B C^{2}$

We know that $D C=\frac{1}{4} B C$

$4 D C=B C$

Substitute $4 D C=B C$ in $A D^{2}=\frac{13}{16} B C^{2}$ we get

$A D^{2}=\frac{13}{16} \times(4 D C)^{2}$

$A D^{2}=\frac{13}{16} \times 16 D C^{2}$

$A D^{2}=\frac{13}{16} \times 16 \times D C^{2}$

$A D^{2}=13 D C^{2}$

Hence we have proved that $A D^{2}=13 D C^{2}$