Let AC be a line segment in the plane and B a point between A and C. Construct isosceles triangles PAB and QBC on one side of the segment AC such that ∠APB = ∠BQC = 120° and an isosceles triangle RAC on the other side of AC such that ∠ARC = 120°. Show that PQR is an equilateral triangle.


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 19293 by Tinkutara last updated on 08/Aug/17

$Let A C be a line segment in the plane$ $and B a point between A and C .$ $Construct isosceles triangles P A B and$ $Q B C on one side of the segment A C$ $such that ∠ A P B = ∠ B Q C = 120 ° and$ $an isosceles triangle R A C on the other$ $side of A C such that ∠ A R C = 120 ° .$ $Show that P Q R is an equilateral$ $triangle .$
Commented by ajfour last updated on 08/Aug/17

Commented by ajfour last updated on 08/Aug/17

${PQ}^{2} = {(a + c)}^{2} + \frac{{(c - a)}^{2}}{3}$ $= \frac{4}{3} (a^{2} + c^{2} + ac)$ ${QR}^{2} = a^{2} + \frac{{(2 c + a)}^{2}}{3}$ $= \frac{4}{3} (a^{2} + c^{2} + ac)$ ${RP}^{2} = c^{2} + \frac{{(2 a + c)}^{2}}{3}$ $= \frac{4}{3} (a^{2} + c^{2} + ac)$ $\Rightarrow PQ = QR = RP, hence △ PQR$ $is equilateral .$
Commented by Tinkutara last updated on 08/Aug/17

$Thank you very much Sir!$
	Answered by Tinkutara last updated on 03/Dec/17

	Commented by Tinkutara last updated on 03/Dec/17

	$T h i s w a s a l s o a s o l u t i o n . I h a v e a d o u b t$ $i n 3 r d l i n e b e l o w t h e f i g u r e . W h y$ $∠ K P Q = ∠ P K L ?$
	Commented by ajfour last updated on 03/Dec/17

	$P Q L K i s a n i s o s c e l e s t r a p e z i u m \sqrt{}$ $I t s b a s e a n g l e s ∠ K P Q i s t h u s$ $e q u a l t o ∠ P K L (w i t h b a s e P K) .$
	Commented by Tinkutara last updated on 03/Dec/17

	$W h y i s o s c e l e s ?$
	Commented by ajfour last updated on 03/Dec/17

	$K i s u p p o s e (i n t h e s o l u t i o n p o s t e d$ $b y y o u) i s r e f l e c t i o n o f P i n A C .$ $L i s r e f l e c t i o n o f Q i n A C .$
	Commented by Tinkutara last updated on 03/Dec/17

	$Thank you Sir!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum