Can You Solve This Plane Geometry Problem Involving an Acute-Angled Triangle?

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SUMMARY

The discussion focuses on solving a plane geometry problem involving an acute-angled triangle PQR, where angle P is π/6. The orthocenter H and midpoint M of QR are utilized to establish relationships between points T, R, and Q, leading to the conclusion that PT equals 2QR. The problem employs properties of parallelograms and isosceles triangles to derive that triangle NPQ is equilateral, confirming the relationship PT = 2PQ.

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  • Understanding of acute-angled triangles and their properties
  • Familiarity with orthocenters and midpoints in geometry
  • Knowledge of parallelogram properties and triangle congruence
  • Basic skills in Euclidean geometry and coordinate geometry
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  • Explore properties of orthocenters in acute-angled triangles
  • Study the relationship between midpoints and triangle segments
  • Learn about parallelogram characteristics in geometric proofs
  • Investigate the implications of equilateral triangles in geometric constructions
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Mathematicians, geometry enthusiasts, and students studying Euclidean geometry who seek to deepen their understanding of triangle properties and geometric proofs.

aniketp
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Hi everyone,
Can anyone solve the followong by plane Euclidean geometry?
I got it by co-ordinate geometry, but couldn't get it by plane...
>In an acute - angled triangle PQR , angle P=\pi/6 , H is the orthocentre, and M is the midpoint of QR . On the line HM , take a point T such that HM=MT. Show that PT=2QR.
 
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Connect points T and R.
Connect points T and Q.
Quadrilateral HRTQ is a parallelogram because HM = MT and MR = MQ.

Thus
TR is perpendicular to PR, and
TQ is perpendicular to PQ.

Let N be the midpoint of PT.
Connect points N and R.
Connect points NQ.
In right triangle PRT, NR = PT/2.
In right triangle PQT, NQ = PT/2

So NR = NQ = PT/2
Thus triangle PQN is isosceles.

Now we will show that < RNQ = 60 degrees.
(note that < NPR = < NRP and < NPQ = < NQP)
< RNQ = < RNT + < QNT
= < NPR + < NRP + < NPQ + < NQP
= 2( < NPR + < NPQ)
= 2 < RPQ
= 2*30
= 60

So triangle NPQ is equilateral
Thus PQ = NQ

As NQ = PT/2
So PQ = PT/2
PT = 2PQ

http://www.idealmath.com
 

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