geometry Q (1 Viewer)

[ToO LaZy ^*]

can't figure it out ..

 

[Eagles]

what were you asked to proof?

 

[ToO LaZy ^*]

ooops..my bad
the diagram shows 2 tangents, AB and AC, drawn from a common point, A, to a circle, centre O

the diameter CE produced cuts the tangent AB at the point D
show that <EDA + 2 x <DEB = 270 degrees

<EDA + 2 x <DEB = 270 degrees

 

[ToO LaZy ^*]

wtf?..it didn't show up properly?!..
show <EDA + 2 x <DEB = 270 degrees

EDIT: can anyone see the bit after 'show', coz it's not showing on mine

 

[Eagles]

nope, but when I quoted it, it turned up

the diameter CE produced cuts the tangent AB at the point D
show that angle EDA + 2 x angle DEB = 270 degrees

EDA + 2 x DEB = 270 degrees

 

[mojako]

Note: make sure you know where point E lies on the diagram as it's invisible there.
Hint: draw the diagram on paper and label alpha, beta, etc... as you read.

Let's construct line EB.
Let's call angle EDA theta, angle DEB alpha.

Let's construct lines BC and BO. call angle CBO beta and angle OBE gamma.
Now pay particular attn to triangle CBD, whose angle sum is
angle CBO + angle OBD + angle BDO + angle OCB = 180 [degrees]
beta + 90 [since AD is a tangent] + theta + beta [since triangle BCO is isosceles] = 180
2*beta + 90 + theta = 180
2*beta + theta = 90 -- (eqn 1)

Now let's utilise our gamma [which is angle OBD]...
angle OBD = angle OBE + angle DBE
90 [since AD is tangent] = gamma + beta [alternate segment, corr. to angle OCB]
--> beta = 90 - gamma

Now let's focus on line DEO...
angle DEB = 180 - angle OEB
--> alpha = 180 - gamma [triangle OBE isosceles] -- (eqn 2)

Let's go back to eqn 1.
2*beta + theta = 90
2*(90 - gamma) + theta = 90
180 - 2*gamma + theta = 90
-2*gamma = - 90 - theta -- (eqn 3)

Now rewrite eqn 2:
2*alpha = 360 - 2*gamma
sub eqn 3: 2*alpha = 360 - 90 - theta
theta + 2*alpha = 270
angle EDA + 2 x angle DEB = 270 degrees

EDIT: in the proof above I didn't use the fact that AC is a tangent. There might be an alternative proof that use that fact, and the fact that AC and AB are from a common external point A.

 

[ToO LaZy ^*]

sorry peeps..
here's the fixed diagram

 

[ToO LaZy ^*]

thanks for the help mojako

 

[CM_Tutor]

I think there's a faster way.

Join CB and BE, and let angle ABC = &alpha;

Now, angle BEC = angle ABC = &alpha; (Alternate Segment Theorem)
And so, angle DEB = 180&deg; - &alpha; _____ (1) (Angle sum of line DEC is 180&deg; )

Also, angle EBC = 90&deg; (Angle in a semicircle is a right angle)
And so, angle EBA = angle EBC + angle CBA = 90&deg; + &alpha;

Now, angle EDB + angle DEB = angle EBA = 90&deg; + &alpha; _____ (2) (Exterior angle theorem, &Delta;DEB)

So, angle EDA + 2 * angle DEB = (angle EDB + angle DEB) + angle DEB, as EDA and EDB are the same angle
= 90&deg; + &alpha; + 180&deg; - &alpha;, using (1) and (2)
= 270&deg;, as required.

 

[ToO LaZy ^*]

oh thx..much faster indeed