Can you say that QR bisects OC
#help 1
amber nimbus
languid flower
yes, what you write is true,
but you may need to prove it
I would go about it this way:
Use the converse of the Intercept theorem (also called Thales theorem or the Side-Splitting theorem for triangles) to show that the lines QR and PS are parallel
(because they are both parallel to BD, the base of two triangles)
Then the same is true for PQ and RS
Now try to prove that angle PQR is right
eternal monolith
Can you say that QR bisects OC
since ∠BCD=90 degrees, QC=CR due to both being half the side lengths of a square. Therefore we may conclude △QCR is a right-angled isosceles triangle, and therefore has two equal base angles, with is 45 degrees due to the sum of inner angles of a triangle si 180 degrees.
Similarly, do the same process for other sides, and we may conclude that all triangles formed by the vertices and mid-points are right angled isosceles triangles, and similarly have 45 degrees base angles.
therefore, ∠PQR = 180-45-45=90 degrees
similarly, for ∠QPS, ∠PSR, ∠SRQ are all right angles
there, with definition of rectangles, we may conclude PQRS is a rectangle
and even more, if you know congruent theorems of triangles, you'll actually find out it's not just a rectangle, but actually a square too!
and yes, QR bisects OC because connect OQ, OR, QR is a diagonal, and triangle congruent theorems proves the proposition.
amber nimbus
Thanks
amber nimbus
since ∠BCD=90 degrees, QC=CR due to both being half the side lengths of a square...
But how can we say that angle BCD is 90 degrees?
And could you please explain me how you used the angle sum property of triangle and stated said that triangle QCR has two 45 degree angles
eternal monolith
But how can we say that angle BCD is 90 degrees?
whoops rhombus sorry, thought ABCD was a square.
then PQRS wouldn't be a square, but still being a rectangle, provable by SAS triangle congruent theorems for PBQ, RSO and APS, QCR, which would still make the four inner angles of PQRS right angles a make it a rectangle.
angle PBQ = angle SDR is given by the definition of a rhombus
eternal monolith
making a pic to show what I exactly mean
amber nimbus
thank you so much @eternal monolith
amber nimbus
sure but i don't know how to do that🥲 i'm a lil new
eternal monolith
say .solved
amber nimbus
.solved
pale spindleBOT
Solved
amber nimbus
need help..
eternal monolith
need help..
simple
just solve the left using distributive property of mulplication
a.k.a time them in
solve it
get x^3+y^3
done
and if you prefer a bit fancier way, you may try understanding how the right became the left, then do the opposite on the left to get right
amber nimbus
i'll try this
proud flicker
which class ncert is this?
amber nimbus
9
proud flicker
Can we say that angle ABC = angle ADC
nope
amber nimbus
ok
amber nimbus
can't we say that they are 45 by proving them equal in some way
proud flicker
only possible case will be each angle is 45
lemme solve it, but i have to presume that BCD is 90
amber nimbus
that's what we need to prove
proud flicker
can't we say that they are 45 by proving them equal in some way
lets try
amber nimbus
sure
proud flicker
that's what we need to prove
thats easy
draw a circle with BD as diameter
diameter subtends 90 at circumference
proud flicker
diameter subtends 90 at circumference
this is a theorem, maybe school wont approve this method
amber nimbus
i think so
proud flicker
got the ans
amber nimbus
ohh noice
