Question

The side AB of a parallelogram ABCD is produced to E such that BE AB  . DE intersects AC and BC at P and Q respectively. Prove that Q is the mid-point of BC . Also prove that 1 PC AP 2  .​

Answer 1

Answer:

$\frac{ {3 > \times { \div 2 { < { \div \div 61 < \frac{ \geqslant |3. = {616. > \sqrt[ - | \div { \\ 3 \frac{ \div .1}{?} \times \frac{?}{?} }^{2} | ]{?} }^{2} | }{?} \times \frac{?}{?} }^{2} }^{?} }^{2} }^{?} }{?}$