According to similarity if
(i) <a =<d,<b=<e, <c = <f
(2) AB/DE = BC/EF = AC/DF
the triangle is similar so....
is it also same as BPT does.this also come under it AQ/AB = AP/AC?!(fig reference)
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There are no points E, F and T? have you posted correctly?
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According to similarity if
<a =<d,<b=<e, <c = <f
Then ∆ABC ~ ∆DEF
After that, we can say
(2) AB/DE = BC/EF = AC/DF
_______________________
In the figure, if PQ is parallel to CB,
then <PAQ = <CAB
<PQA = <CBA
<APQ = <ACB
So ∆AQP ~ ∆ABC
Now, we can say AQ/AB = AP/AC
<a =<d,<b=<e, <c = <f
Then ∆ABC ~ ∆DEF
After that, we can say
(2) AB/DE = BC/EF = AC/DF
_______________________
In the figure, if PQ is parallel to CB,
then <PAQ = <CAB
<PQA = <CBA
<APQ = <ACB
So ∆AQP ~ ∆ABC
Now, we can say AQ/AB = AP/AC
if PQ is not parallel to CB, then it wont be true.
so after proving the angles equal thn only we can say that the ratios of their side is equal?! huh
so we consider the ratio of AQ/AB similar not AQ directly equal to AB
that's one way. You need to prove the triangles are similar. you can use any condition. i have used AAA similarity condition.
ohkay.....
you can use SSS similarity condition or any other, i dont remember all
so even if i prove the ratio of sides are equal thn also i can say that the angles are equal after proving them similar ?! huh
similarity and congruency is in 7th class(in cbse). you must know it
yes..
ohkay got it..
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