Question

pls hep me with question

ABCD is a parallelogram and P and R are the mid points of side DC and BC respectively . if line PR intersect daignol AC at Q , prove that AC = 4CQ . no absurd answers please

Answer 1

$\huge{\pink{\underline{\ulcorner{\star\: Solution\: \star}\urcorner}}}$

Given : ABCD is a parallelogram and DC have a mid point P and BC have R

Diagonal AC and BD bisect each other at point O

Construction : Join mid points P and R which cut the diagonal AC at Q

To Prove : $AC = 4\times CQ$

Proof :

We have a diagonal AC

$OC = \frac{1}{2} AC$

Statement 1 : Diagonal of Parallelogram bisect each other in equal parts.

Construction 2 : Join the OP & OR parallel to BC and DC which form a parallelogram

That's mean PC = OR & OP = RC (side of Parallelogram opposite to each other are equal)

Now PQ = QR & OQ = QC (refer statement 1)

means

$QC=\frac{1}{2} OC$....(1)

AC = AO + OC

AO = OC( refer statement 1 )

$AC= 2OC \\\implies AC = 2(2QC) \\\implies AC = 4QC. .....(by\: 1) \\\implies AC = 4CQ$

$\red{\underline{Hence \: Proved}}$

$\orange{\boxed{AC = 4CQ}}$