202 views

0 votes

In a triangle $\text{ABC, AB= 6, BC= 8}$ and $\text{AC=10}.$ A perpendicular dropped from $\text{B}$, meets the side $\text{AC}$ at $\text{D}.$ A circles of radius $\text{BD}$ (with centre $\text{B})$ is drawn. If the circle cuts $\text{AB}$ and $\text{BC}$ at $\text{P}$ and $\text{Q}$ respectively, then $\text{AP : QC}$ is equal to

- $1:1$
- $3:2$
- $4:1$
- $3:8$