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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).

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what's the meaning of ar($\triangle$APB)+ar($\triangle$BQC)+ar($\triangle$ASC)? –  lab bhattacharjee Sep 23 '12 at 14:18
    
Im guessing that its three external angles based on the respective sides, but i could be wrong... These people are paid to make confusing questions D: –  Aayush Agrawal Sep 23 '12 at 14:19
    
Could you tell me what's to be proven? –  lab bhattacharjee Sep 23 '12 at 14:21
    
Prove that ar(△APB)+ar(△BQC)=ar(△ASC) –  Aayush Agrawal Sep 23 '12 at 14:26
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3 Answers 3

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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 $|AB|^2+|BC|^2=|CA|^2$

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Thx, i just added a second question too, just realised i forgot to add that earlier. Can you take a loot at that too? Thx alot! –  Aayush Agrawal Sep 23 '12 at 14:16
    
@AndréNicolas, thanks for your observation, I've rectified now. –  lab bhattacharjee Sep 23 '12 at 15:08
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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)$

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Thx man, i just realised i forgot to add a second question :( can you take a look at that one too –  Aayush Agrawal Sep 23 '12 at 14:16
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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.

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