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

[aniketp]

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.

 

[redmafiya]

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

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