Question

please solve.....❤️............

Answer 1

Hello Saloni.

Thanks for asking this question.

Your answer is =>

In the given figure , taking two different triangles $\triangle ABD$ & $\triangle CBD$ under observation.

in these triangles ,

side CD || side AB

SO ,

$\angle CDB = \angle ABD$............(1)

also , side AD || side BC

so ,

$\angle ADB = \angle DBC$...........(2)

from eq. (1) & (2) ,

By A-A ( angle - angle ) test ,

$\triangle ABD = \triangle BCD$

so , now we can say that ,

$\frac{AO}{CO} = \frac{CO}{AO}$

but , P and Q are the midpoints of side CD & side CB and their joining line PQ intersects diagonal AC.

so , it is clear that , segment PQ bisects the half seg.CO of diagonal AC.

=> CO = 2 × CR

also ,

=> CO = $\frac{1}{2} \times AC$

also , By A-A test , we can prove that ,

$\triangle CPR = \triangle CQR$

=> seg. PC = seg. CQ = seg. CR

=> seg. CO = 2 × CR......(3)

now we have already proved that ,

$\frac{AO}{CO} = \frac{CO}{AO}$

multiply both sides by 2.

$2 \times \frac{AO}{CO} = 2 \times \frac{CO}{AO}$

BUT ,

=> CO = AO

also , we know that ,

=> 2 × AO = AC

and ,

=> CO = 2 × AR

=> 2CO = 2(2 CR ) = 4 CR

SO , Our equation becomes ,

$=> \frac{AC}{\cancel{CO}} = \frac{4 \times CR}{\cancel{CO}}$

$=> AC \:=\:4 \times CR$

so ,

$=> \boxed {CR = \frac{1}{4}\:AC}$



$\huge \bf { \# \mathbb {B}e \mathbb{B}rainly}$