AD + CE = AE, computational inside a right triangle (Singapore Junior 2008)

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AD + CE = AE, computational inside a right triangle (Singapore Junior 2008)

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#1 h Jul 10, 2019, 1:25 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

In $\vartriangle ABC, \angle ACB = 90^o, D$ is the foot of the altitude from $C$ to $AB$ and $E$ is the point on the side $BC$ such that $CE = BD/2$ . Prove that $AD + CE = AE$ .

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#2 h Jul 10, 2019, 2:02 PM • 5 Y

Y by Path_to_Almighty, RavenKing34, ImSh95, Adventure10, Mango247

Solution

This post has been edited 1 time. Last edited by amar_04, Jul 10, 2019, 2:03 PM

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#3 h Jul 10, 2019, 2:04 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Synthetic Hint?

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#4 h Jun 29, 2023, 5:54 PM • 1 Y

Y by ImSh95

Let AD=x
AND LET <DCB=THETA
AND CE=Y DB=2Y BY ABOVE CONDITION
Then apply sine rule in triangle adc and dcb
CD/SIN(90-THETA)=2Y/SINTHETA
CD/SINTHETA=x/SIN(90-THETA)
DIVIDE THOSE TWO EQUATION YOU WILL GET 2Y/X=1 AND X=2Y
THEN LET THE <CAE=ALPHA SO SIDE AC =YCOT(ALPHA)
SO SIDE AE^2=Y^2(COT^2(ALPHA)+1)
BUT YOU CAN APPLY AGAIN SINE RULE IN TRAINGLES ACE AND AEB
SIN(90+ALPHA)/4Y=SIN45/AE
SIN(ALPHA)/Y=SIN(90)/AE
YOU WILL GET COT(ALPHA)=4/√2 SO PUT THIS VALUE IN THE VALUE OF AE^2=9y^2
Or AE=3Y BUT YOU CAN CLEARLY SEE THAT AD+CE=3Y=AE

HENCE PROVED!!

PLS DONT JUDGE ME AS IAM NEW TO HERE NOT VERY GOOD AT TYPING PLS IGNORE THE TYPING MISTAKE AS IAM JUST 14 YEARS OLD DUMB WHO KNOWS NOTHING

This post has been edited 1 time. Last edited by Mathgloggers, Jun 29, 2023, 5:55 PM
Reason: Typing error

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#5 h Jun 30, 2023, 2:37 PM • 1 Y

Y by ImSh95

parmenides51 wrote:

In $\vartriangle ABC, \angle ACB = 90^o, D$ is the foot of the altitude from $C$ to $AB$ and $E$ is the point on the side $BC$ such that $CE = BD/2$ . Prove that $AD + CE = AE$ .

If $M$ is the midpoint of $DB$ and $CE=DM=MB=x,$ I'll prove that $AE=c-x.$

$A{E^2} = {b^2} + {x^2} = {c^2} - {a^2} + {x^2} = {c^2} - BD \cdot BA + {x^2} = {c^2} - 2cx + {x^2} = {(c - x)^2},$ done.

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