Geometry Problem 1487. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

## Saturday, November 28, 2020

### Geometry Problem 1487: Right Triangle, Altitude, Angle Bisectors, Measurement

Labels:
altitude,
angle bisector,
geometry problem,
measurement,
right angle

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https://photos.app.goo.gl/aXxd5z7tmrZBAcbA8

ReplyDeleteDraw incircle of triangle ABC with radius r

CD meet BE at K ( see sketch)

Calculate AC= 10 , perimeter 2p= 24 , area (ABC)= 24 , r= S/p= 2

We have ∠ (ADC)=135 and ∠ (MDK)=45; ∠ (ABH)= ∠ (ACB)

∠ (BMD)= ½(∠ (BAH)+ ∠ (ABH))= 45 => CD ⊥BE

So triangle ECB is isosceles and DE=DB= r.sqrt(2)= 2.sqrt(2)

AB²= AH*AC or 36 =AH*10 which yields AH = 18/5 while BH = (AB*AC)/BC = 24/5. AE/EH = AB/BH or AE/AH = AB/(AB+BH) Plugging in the values as above, AH=2 and thus EC =8/ We take B as the origin and BA & BC as x-axis & y-axis resply. Then, by the division formula, E:(24/5,8/5) while D: (2.2) since the in-radius'r'of Tr. ABC =(8+6-10)/2 = 2, Finally, DE² = (24/5-2)²+ (8/5-2)² = 8 or DE = 2√2

ReplyDeleteED=2V2 (V = sqrt)

ReplyDeleteSeee sketch for deatails

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Deletehttps://photos.app.goo.gl/TvawRXPuoCnVamtw7

DeletePlease post the the third

https://photos.app.goo.gl/ZGQHdBkPirQBti2Y6

Delete< DAE = 45 - C/2 = < EBD, hence ABDE is concyclic and ED = DB = y say

ReplyDeleteNow < DBC = < DBA = < DEC = 45 hence Tr.s BCD & ECD are congruent ASA

and so CE = CB = 8, whence AE = 2

Therefore EH = 6^2/10 - 2 = 8/5 & since BH = 6 X 8 / 10 = 24/5,

BE^2 = (24/5)^2 + (8/5)^2 ==> BE = (8/5).sqrt10

Now since < ADE = < ABE = < ACD = C/2, AD^2 = 2 X 10 and so AD = 2.sqrt5

Applying Ptolemy in cyclic quad ABDE, 6y + 2y = BE.AD = (8/5)sqrt10 X 2.sqrt5

Simplifying y = 2.sqrt2

Sumith Peiris

Moratuwa

Sri Lanka

Let Perpendicular from D meet AC at P

ReplyDeleteIncircle Radius DP of ABC = 2

Incircle Radius r1 of ABH = 1.2

Incircle Radius r2 of ABC = 1.6

HP=(r2-r1)=0.4 , AH=3.6, HB=4.8 , EH=1.6 (EH=HB.AE/AB)

Apply pythogorus to right-isoscelse EDP=> ED=2Sqrt(2)

Why HP= r2-r1 ? Please explain

DeletePeter Tran

Considering using triangle notations

ReplyDeletem(ABD)=45=> m(EBD)=45-C+C/2=45-C/2=A/2=m(DAE)

So BDEA is concyclic.

m(BED)=A/2 and m(BEH)=A+C/2=45+A/2

=>m(DEH)=m(BEH)-m(BED)=45

In-circle Radius of ABC = 2 [using r.(a+b+c)=a.c]

Let P be the foot of Perpendicular from D to AC

It can be easily found that DEP is right isosceles => DE=2Sqrt2

ITS A GOOD GEOMETRY

ReplyDelete