Triangle $\mathrm{ABC}$ is similar to triangle $\mathrm{PQR}.$ If $\mathrm{AD}$ and $\mathrm{PM}$ are altitudes of the two triangles, prove that : $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}.$

 

In the following diagram, lines $\mathrm{l}, \mathrm{m}$ and $\mathrm{n}$ are parallel to each other. Two transversals $\mathrm{p}$ and $\mathrm{q}$ intersect the parallel lines at points $\mathrm{A}, \mathrm{B},\mathrm{C}$ and $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ as shown.

Prove that : $\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{PQ}}{\mathrm{QR}}$
 

In the following figure, $\mathrm{DE}\parallel \mathrm{AC}$ and $\mathrm{DC}\parallel$$\mathrm{AP}.$ prove that: $\frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BC}}{\mathrm{CP}}.$

In the figure given below, $\mathrm{AB}\parallel \mathrm{EF}\parallel \mathrm{CD}.$ If $\mathrm{AB} = 22.5 \mathrm{cm}, \mathrm{EP} = 7.5 \mathrm{cm}, \mathrm{PC} = 15 \mathrm{cm}$ and $\mathrm{DC} = 27 \mathrm{cm}.$ Calculate: 

$\mathrm{EF}$ and $\mathrm{AC}$

In $△\mathrm{ABC}, \angle \mathrm{ABC} = \angle \mathrm{DAC}, \mathrm{AB} = 8 \mathrm{cm}, \mathrm{AC} = 4 \mathrm{cm}$ and $\mathrm{AD} = 5 \mathrm{cm}.$

Prove that $△\mathrm{ACD}$ is similar to $△\mathrm{BCA}.$

In $△\mathrm{ABC}, \angle \mathrm{ABC} = \angle \mathrm{DAC}, \mathrm{AB} = 8 \mathrm{cm}, \mathrm{AC} = 4 \mathrm{cm}$ and $\mathrm{AD} = 5 \mathrm{cm}.$

Find $\mathrm{BC}$ and $\mathrm{CD}.$

In $△\mathrm{ABC}, \angle \mathrm{ABC} = \angle \mathrm{DAC}, \mathrm{AB} = 8 \mathrm{cm}, \mathrm{AC} = 4 \mathrm{cm}$ and $\mathrm{AD} = 5 \mathrm{cm}.$

Find area of $△\mathrm{ACD}:$ area of $△\mathrm{ABC}.$