#Euclidean Geometry (∆)

1 messages · Page 1 of 1 (latest)

tardy karma
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A triangle - rectangle problem

The side AB in the rectangle ABCD is twice the side BC. A point P is
taken on the side AB so that BP =
4 AB. Show that BD is perpendicular to
CP.

bitter oracleBOT
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bitter oracleBOT
vernal sand
tardy karma
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I uhh

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Constructed to Bisect AB in Q and drew QR J AB to meet BD in

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QR perpendicular to AB to meet BD inR

vernal sand
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That all feels a little arbitrary.

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You want to know what I think?

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@tardy karma?

vernal sand
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I spotted what might be a mistake.

tardy karma
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Hmm om

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Ok

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Wait...

vernal sand
tardy karma
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Yeah here is it

image0.jpg
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It's ¼AB=BP not BP=4AB

vernal sand
# tardy karma It's ¼AB=BP not BP=4AB

Right, that's what I thought the typo was, except I thought you meant that BP = 4 AP, because for BP = 4 AB, you would need P to be on the extension of AB, not on AB.

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Now, onto actually solving the problem. We're given a bunch of length ratios, and then asked to prove that two line segments are perpendicular.

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So what theorems do we have handy that relate lengths to right angles?

tardy karma
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If one side of a triangle is produced, (i) the exterior angle is equal to the
sum of the interior non-adjacent angles; (ii) the sum of the three angles of a
triangle is two right angles.
COROLLARY 1. If two angles of one triangle are respectively equal to two
angles of another triangle, the third angles are equal and the triangles are
equiangular.
COROLLARY 2. If one side of a triangle is produced, the exterior angle is
greater than either of the interior non-adjacent angles.

vernal sand
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Um.

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I was thinking of the Pythagorean theorem.

tardy karma
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Oh I thought u needed em

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Sry

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Mb

tardy karma
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Ok how do you plan to solve the problem

vernal sand
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Well, if we draw out the diagram and call the intersection of BD and CP Q, then we wish merely to prove that one of the - wait.

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No, I thought I might've spotted an easier method involving quadrilaterals.

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Or, wait.

tardy karma
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Wait I see somthn

vernal sand
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To clarify, my idea is to prove that one of the triangles containing point Q is right by proving its lengths satisfy the Pythagorean theorem.

tardy karma
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AB = 2 BC, .'. BC = BQ. In the ∆ABD, Q is the mid-
point of AB and QR =¼ AD. .'. R is the mid-point of BD and QR
= ½AD (

vernal sand
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Wait, prove that?

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What's R here now?

tardy karma
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I drew it

vernal sand
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Where are you getting BC = BQ?

tardy karma
tardy karma
vernal sand
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Are we talking about the same Q?

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So wait, R is the midpoint of AB?

tardy karma
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Yeah

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Acc to my constñ

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So BC=BQ

vernal sand
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No, BC = BR.

tardy karma
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Hiw

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How!?

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How*

vernal sand
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...because you just said R is the midpoint of AB.

tardy karma
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Now aait

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Lemme make a figure

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And send u

vernal sand
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So AB = 2 BC ==> BC = AB/2, and AR = BR = AB/2 by the definition of midpoint.

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Hence BC = BR.

tardy karma
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See my fig

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I hope the confusion is clear

vernal sand
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If BC = BQ, that makes triangle BQC isosceles, with B as the vertex, therefore <BCQ = <BQC, which makes it impossible for <BQC to be right, which is what we're trying to prove it is.

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Okay, you misspoke.

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The point you're calling F is the point I'm calling Q.

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And the point you've labeled as Q is the point you told me was R.

tardy karma
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No Ig it was a typo mb

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Yeah so waht I propose is...

vernal sand
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I see. QR is by construction the perpendicular bisector of AB.

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Therefore triangle BCP is congruent to triangle QBR.

tardy karma
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AB = 2 BC, .'. BC = BQ. In the triangle ABD, Q is the mid-
point of AB and QR = ½ AD. .'. R is the mid-point of BD and QR
=½ AD (

vernal sand
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Okay, I'm sorry, but you're really leaving out a lot of important details.

tardy karma
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Like...!?

vernal sand
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It makes no sense to write one thing and then write "therefore".

tardy karma
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Wdym

vernal sand
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AB = 2 BC, therefore BC = BQ makes no sense.

tardy karma
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Bro BQ=½AB

vernal sand
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But you didn't say that.

tardy karma
vernal sand
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What you need to say is AB = 2 BC is given, therefore BC = AB/2. Q is the midpoint of AB by construction, therefore BQ = AB/2 by the definition of a midpoint. Therefore, BC = BQ (by transitivity of equality).

tardy karma
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Exactly

vernal sand
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Yes, that's exactly what you should've said and didn't.

tardy karma
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Bro I'm on ma phone how can I type so much 😭

vernal sand
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I mean, it's called patience, and you can't do math without it.

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You have to cross your t's and dot your i's or else your proof is just wrong.

tardy karma
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Ok 😭

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Im sorry bruh

vernal sand
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Don't apologize to me, sloppy thinking is a habit that hurts you.

tardy karma
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Okie 👍

vernal sand
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Now, QR = AD/2, prove that.

tardy karma
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Similar ∆s

vernal sand
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Which triangles? Prove they're similar.

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Also, how does that result in the equation?

tardy karma
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∆BQR&∆ADB

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angle b=angle b and BQR=BAD=90°

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So...

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BQ/BA = QR/AD

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I.e. 1/2

vernal sand
tardy karma
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It's my constñ

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😑

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Bro...

vernal sand
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Okay, look.

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BAD = 90 degrees because ABCD is a rectangle.

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BQR = 90 degrees by construction because QR is by construction perpendicular to AB.

tardy karma
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So...!?

vernal sand
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Therefore BAD = BQR.

tardy karma
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That is waht I'm saying

vernal sand
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No, it's what you're not saying.

tardy karma
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Bruh

vernal sand
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That's the whole problem.

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It's what you're assuming, but not stating.

tardy karma
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I am impatient 😭

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Oh I solved the problem

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And I'm gonna tell you how with patient this time so wait ...

vernal sand
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Oh wait! Triangle DCB is similar to triangle CBP.

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Which I think can then be leveraged into proving that triangle BFC is similar to triangle PBC.

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Got it!

tardy karma
vernal sand
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...you exported it to a pdf?

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Why not a plaintext file?

tardy karma
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Why do you inqire so much 😭

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Bud I'm on ma phone and it's LateX

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And I couldn't do it here cuz my text was too long

vernal sand
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Well, pdfs are kind of sketchy last I heard.

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In any case, I figured it out.

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Look, just explain each step, one at a time, in plain words.

dusky sparrow
# tardy karma

ur picture looks likeAB=4BP，your topic is the opposite！According to your picture.
AssumeBP=x,soAB=4x,BC=2x=AD

It can be seen from the rectangle.
CP=5^(1/2)x ,BD=2(5)^1/2
BD/CP=BA/CB=BP/DA Triangles are similar.so AngleABD=AngleBCP becauseAngleBCP+BPC=90 AnglePBD+BPC=90,AngleBFP=180-90=90 BD is perpendicular to CP

tardy karma
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Anyways I solved the problem