A cute question - Coordinate geometry
Hmm.. I'm doing this now
that's not cute,looks fearsome, jokin...
If there is a cute solution, then this would be a cute question :)
is BG perpendicular to AC???
if it is, then i think i may have something...
(a)(i) can be done by simply stating: AngleBAG=AngleBDG AngleABG=AngleGBD(Incentre is the point of intersection of angle bisectors, so BI bisects angle ABD)
if I is an incenter that means BG is an angle bisector...
@Sarkar, I don't think so. Looks can be deceiving!
that's my line...
so SAS , they're congruent.
This is not a drama! Is my solution clear?
crystal
exactly @dpalnc
So obiviouly Bg is perpendicular to aG.
@dpaInc lol My workings for ai: In ΔABG and Δ DBG ∠ABG = ∠ DBC (How should I write the reason?) BA = BD ( given) ∠BAG = ∠BDG (base ∠, isosΔ) ∴ ΔABG (is congruent to) Δ DBG
*AC
Triangle AI ANd Triangle ABE Angle ABE = Angle AGI Angle BAI = IAG => SImilar triangles Proves b
"ABG = ∠ DBC (How should I write the reason?)" i think you mean angle DBG?
Sorry.. yup...
BG is an angle bisector since BG goes thru the incenter I
Yup.. But how should I present it? Should it be like this? since BG goes through the incenter of the triangle, BG is the angle bisector, that is ∠ ABG = ∠ DBG In ΔABG and Δ DBG ∠ABG = ∠ DBG (proved) BA = BD ( given) ∠BAG = ∠BDG (base ∠, isosΔ) ∴ ΔABG (is congruent to) Δ DBG
^^ Seems excellent presentation.
i see you're using ASA. that looks fine too.
Oh... I forgot to write ASA ... sorry
@dpaInc another one is AAS, where S is BG=BG ?
Hmm.. G= (-7,0) ?
Yes.
Simple midpoint.
Ad Equation of Circle: \[(x+7)^2 + (y - 9)^2 = 405\]
I= (-7, 9)?
Yep.
Radii = 9sqrt(5)
Good. I got them all. This one is not a cute one :(
Haha. It's easy. Justl mixed jumbled up to confuse you. Real Charmer.
The key word is incentre... This troubles me :( I'm not familiar with the centres in a circle
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