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 $|AB|^2+|BC|^2=|CA|^2$
A: Part 1,
In, $\Delta ABD, \frac{z}{x}=\frac{QD}{BD}$
In, $\Delta BCD, \frac{z}{y}=\frac{BQ}{BD}$
These equations $\implies \frac{z}{x}+\frac{z}{y}=\frac{BQ+QD}{BD}=1\implies \frac{1}{x}+\frac{1}{y}=\frac{1}{z}$
Part 2,
ar($\triangle$APB)+ar($\triangle$BQC)=$\frac{\sqrt 3}{4}(|AB|^2+|BC|^2)=\frac{\sqrt 3}{4}|AC|^2=$ar($\triangle$ASC)$
A: For the added question, note that the area of an equilateral triangle of side $a$ is $\lambda a^2$ for some constant $\lambda$. By using properties of special angles, or otherwise, it can be shown that actually $\lambda=\frac{\sqrt{3}}{4}$, but that is not really needed here. For details, draw the height from some vertex, and note that the height is $a\sin(60^\circ)$.
From the Pythagorean Theorem, we have $(BA)^2+(BC)^2=(AC)^2$. Multiplying through by $\lambda$ we get $\lambda(BA)^2+\lambda(BC)^2=\lambda(AC)^2$, which gives the desired area result.
Remark: Adding to a question is not a good idea. This should have been an entirely new question. 
