See complete Problem 132

Triangle, Angle,60, Orthocenter, Congruence, Midpoint. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Saturday, June 28, 2008

### Elearn Geometry Problem 132

Labels:
30-60,
angle,
congruence,
midpoint,
orthocenter,
triangle

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To prove the proposed theorem, two steps are needed.

ReplyDeleteFirstly, prove that EH is half of BD (E is the midpoint of BD).

Secondly, prove that the triangle AHE is equilateral. It follows that AH = EH = (1/2)BD.

The details are as follows:

Connect A and D, D and H.

Extend CA to F.

Then ADHC is a parallelogram(as M is the midpoint of both line segments of AH and CD).

mDAB is a right angle(as mDAF = mHCA = mABH = 30 degrees, and mBAC = 60 degrees).

mDHB = 90 degrees (as DH is perpendicular to AC).

In right triangle DAB, AE = (1/2)BD.

In right triangle DHB, EH = (1/2)BD.

So triangle AEH is an isosceles triangle.

Also note that quadrilateral DAHB is cyclic (as mDAB and mDHB are right angles), we have:

mEAB = mEBA = mDHA = mHAC.

So mEAH = mEAB + mBAH = mHAC + mBAH = mBAC = 60 degrees.

So isosceles triangle EAH is equilateral.

Let CH cut BA at X, BH cut AC at Y, BA cut DH at Z

ReplyDeleteACHD is a //gram, so mDAB=mAXC=mBYA=mBHD

A,D,B,H concyclic and ZHA similar to ZBD

HA/BD=ZH/ZB=AY/AB=1/2

Let AH intersect BC in I.

ReplyDeleteADHC is a //gram so AD//HC, ΔABD is right in A and AD=HC.

In right triangles ΔIAB, ΔICH, ΔABD and ΔIAH:

AB=2xAI and HC=2xIH=AD hence ΔABD and ΔIAH are SAS similar and therefore BD=2xAH QED.

Amend first sentence to read : "Let BH intersect AC in I."

DeleteMany thanks to rv.littleman for spotting the mistake.

Let E be the mid-point of BC

ReplyDeleteJoin ME and since M is the mid-point of DC=> ME=BD/2

Let F be the foot of the perpendicular from C to AB

Note that M,E and F lie on the nine-point circle of ABC and with simple angle chasing it can be deduced that Tr.MEF is a 60-30-90 triangle (Proof below) . Hence MF=ME/2

But given M is the mid-point of AH and since m(AFC)=90=m(AFH)

=> AM=MH=MF (M circum-center of Right Tr. AFH)

Thus AH=ME=BD/2 Q.E.D

Additional notes to prove MEF a 30-60-90 Tr.

Let G be foot of Perpendicular from A to BC

AFGC is cyclic & given m(A)=60=> m(FGC)=120

=>m(BGF)=60 -------(1)

& m(FGA)=30 as m(AGC)=90 --------(2)

As F,M,G & E lie on nine-point circle of ABC, m(MEF)=m(FGA)=30

and m(FME)=m(BGF)=60 => MEF is a 60-30-90 Tr.

Let HD,AB intersect at P

ReplyDeleteACHD is a parallelogram so < ABH = < ACH = < ADP

Hence ADBH is concyclic and Tr.s BDP and APH are similar

This gives PH = BP/2 therefore AH = BD/2

Sumith Peiris

Moratuwa

Sri Lanka