Q:

Three geometry questions

In figure AB$\parallel PQ \parallel CD$ Prove that $\frac 1 x + \frac 1 y =  \frac 1 z$
Equilateral triangles APB, BQC and ASC are described on each side of a right-angled 
triangle ABC, right angled at B.  Then prove that ar($\triangle$APB)+ar($\triangle$BQC)=ar($\triangle$ASC).

A:

As $\triangle ABD,\triangle PQD$ are similar as $\angle ABD=\angle PQD$ ,so $$\frac z x= \frac{QD}{BD} $$
Similarly, as $\triangle BCD,\triangle BPQ$ are similar ,so $$\frac  z y = \frac{BQ}{BD} $$
So, $$\frac z x+\frac  z y=\frac{QD}{BD}+ \frac{BQ}{BD}=1$$
The area of $\triangle APB=\frac{\sqrt3}2|AB|^2$
So, the area$(\triangle APB)$+ area $(\triangle BQC)$
$=\frac{\sqrt3}4(|AB|^2+|BC|^2)=\frac{\sqrt3}4 |CA|^2=$area$(\triangle ASC)$ as $|AB|^2+|BC|^2=|CA|^2$

