Question

Please solve this Please give the correct answer, do not response if you don't know ​​

Answer 1

Answer:

Step-by-step explanation:

in ∆ ABC , AB || EF

∴ ∆ ABC is similar to ∆ EFC   [ AA similarity crirerion ]

∴ $\frac{AB}{EF} = \frac{BC}{FC} => \frac{a}{h} = \frac{p}{FC}$

∴ $FC = \frac{ph}{a}$ ..........(1)

in ∆ BCD, EF || DC

∴ ∆ DCB is similar to ∆ EFB [ AA similarity crirerion ]

∴ $\frac{DC}{EF} = \frac{BC}{FB} => \frac{b}{h} = \frac{p}{FB}$

∴$FB = \frac{ph}{b}$ ...........(2)

Adding (1) and (2)

FC + FB = $\frac{ph}{a} +\frac{ph}{b}$

∴$p = ph [\frac{a+b}{ab} ]$

∴  $h = \frac{ab}{a+b}$

Answer 2

Answer:

hyy here is your answer