Proposed Problem

Click the figure below to see the complete problem 363 about Right triangle, Congruence, and Angles.

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Complete Geometry Problem 363

Level: High School, SAT Prep, College geometry

## Saturday, October 10, 2009

### Problem 363: Right triangle, Congruence, Angles

Labels:
angle,
congruence,
right triangle

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By considering isos triangle, alpha=45

ReplyDeleteThen, by considering tan(beta+theta),

tan(beta+theta)=(tan(beta)+tan(theta))/(1+tan(beta)tan(theta))

=(1/2+1/3)/(1-1/2*1/3)

=1

So beta+theta=45,

alpha+beta+theta=90

How do you solve the problem without useing trigonometric approch???

ReplyDeleteJust useing pure Geometry!!!!

Is there anybody who has any idea???

Let D_1 be a point with ABDD_1 is a square and let D_2 be a point with DD_1 =DD_2 and BD is a perpendcular to D_1D_2.

ReplyDeleteThen we have D_2A=D_2C and ang(D_2CD)=ang(AEB)=beta, that is, the triangle D_2AC is a right isoscele triangle. Therfore ang(D_2CA)=45=beta +gamma and alpha=45. We get

ang(ADB)+ang(AEB)+ang(ACB)=alpha+beta+gamma=90 degrees.

sum of the angles ADB, AEB, and ACB is 90 degrees.

BDer

1.in tri ABD, ang.ADB=45 ...(iso. tri thm)

ReplyDelete2.hence alpha=45degree

3.tri. DEA similar to tri CAD ...(sss test of simlarity)

4.hence ang.AED=ang.DAC ...(c.a.s.t)

5.hence ang.DAC=Beta ...(From 4)

6.in triangle ADC,

ang.DAC+ang.ACD=ang.ADB...(exterior ang. thm)

7.hence beta+theta=45 degree...(from 5,6 & given)

8.hence alpha+beta+theta=45+45=90 degree ...(from 2 & 7)

Q.E.D

Since alpha=45 we need to prove that beta + theta = 45.

ReplyDeleteThis can be seen by noting that AD^2=DE.DC. Hence AD is tangential to circle AEC and so < DAE = theta = 45 - beta. QED

Sumith Peiris

Moratuwa

Sri Lanka

Let AB=1.

ReplyDeleteSince ABC is isosceles, alpha = 45 degrees.

tan beta=1/2 and tan theta = 1/3

so beta = arctan 1/2 and theta = arctan 1/3

Therefore,

45 + arctan 1/2 + arctan 1/3 = 90 degrees

Make ABDF a square and note that since ΔADE and ΔCDA are similar (SAS), <DAC = <DEA = β. Therefore <BAD = α, <DAC = β and <CAF = θ, hence α+β+θ =90°.

ReplyDeleteAnother Geometry Proof

ReplyDeleteComplete squares ABDX, XDEY, AXWZ

In Tr.s ABE & AWC, BE/AB = CW/AW = 2. Further < ABE = AWC = 90

It follows that the 2 Tr.s are similar & so < AEB = < ACW i.e Beta = 45 - Theta

Hence Beta + Theta = 45 and since Alpha = 45 the result follows

Sumith Peiris

Moratuwa

Sri Lanka

See the drawing: Drawing

ReplyDeleteBA=BD and α = 45°

Pythagoras : AD^2=2, AE^2=5, AC^2=10 => AC^2= 2AE^2

This means that AC is the hypothenuse of a square AE x AE

Inverse the drawing and translate it on the left of a distance BD

Define G and F respectively the equivalents of B and A

α = 45° => ΔABF is congruent to ΔDBF (SAS)

=> AF=DF

DF=AE by construction => AF=AE

Therefore ΔFAE is isosceles, and since AC is the hypothenuse of a square AE x AE => β+θ =45° => α+β+θ =90°