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I am currently studying euclidean geometry and there's is a problem that really boggles me. Actually, even knowing the answer I cannot figure how to "draw" the problem. Here it is:

$ABC$ is a triangle right-angled at $A$: and the sides $AB, A$C are intersected by a straight line $PQ$, and $BQ, PC$ are joined. Prove that

$$ BQ^2 + PC^2 = BC^2 + PQ^2$$

Here is the answer:

$$BQ^2 = BA^2 + AQ^2, CP^2 = CA^2 + AP^2$$

adding $$BQ^2 + CP^2 = (AB^2 + AC^2) + (AQ^2 + AP^2) = BC^2 + PQ^2$$

Reading the problem, I imagine a right-angled triangle where a straight line cross $AB$ & $AC$ but it doesn't make any sense with the solution &/or the thing we're supposed to show.

If anyone could help me understand the problem it would be great ! I'm quite confident with the pythagoras theorem in itself, more worried about understanding the problem. Thanks for your time.

Siong Thye Goh
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le sram du bled
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1 Answers1

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The solution considers three different right angled triangles as shown in the figure below:

enter image description here The triangles at stake are: $ABC$, $APC$, and $ABQ$.

zoli
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