## Saturday, October 10, 2009

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

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

See more:
Complete Geometry Problem 363
Level: High School, SAT Prep, College geometry

1. By considering isos triangle, alpha=45
Then, 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

2. How do you solve the problem without useing trigonometric approch???
Just useing pure Geometry!!!!
Is there anybody who has any idea???

3. bae deok rak(bdr@korea.com)October 2, 2010 at 5:43 PM

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.
Then 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

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

BDer

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

2.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)

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

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

6. Let AB=1.
Since 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

7. 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°.

8. Another Geometry Proof

Complete 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

9. See the drawing: Drawing

BA=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°