Question

In ∆ABC, AB=AC. P is the midpoint of AC and Q is the midpoint of AB. Prove that quadrilateral BCPQ is cyclic.​

Answer 1

$\\ \\ Given :  \\ \\ ABC is \: a \: triangle\\ \\ \\  PQ∣∣BC \\ \\ AD  \: is \: the \: median \\ which \: cuts \:  PQ  \: at \: R \: \: \\ \\ \\ </p><p>To \: prove : \\ \\ \: AD bisects PQ at R. \\ \\ \\ Proof : \\ \\ In ΔABD \\ \\ PR∣∣BD \\ \\ </p><p></p><p> \frac{AP}{PB} = \: \frac{AR}{RD} \: \: \: ( \: BPT \: ) \\ \\ \\ </p><p></p><p>In ΔACD \\ \\ RQ∣∣DC \\ \\ \frac{AR}{RD} = \frac{AQ}{RD} ( \: BPT \: ) \\ \\ \\ In  \: ΔAPR  \: and \:  ΔABD,  \\ \\ \\ ∠APR=∠ABD \\  (corresponding \: angles) \\ \\ </p><p></p><p>∠ARP=∠ADB \\  (corresponding \: angles) \\ \\ \\ </p><p>∴ΔAPR  \: is \: similar \: to \:  ΔABD \\ \:  ( \: AA \: similarity \: ) \\ \\ \\ ∴AP=AR=PR \\  (corresponding \: sides \: of \: \\ similar \: triangles \\ \: are \: proportinal \: ) - - - eq(i)\\ \\ \\ AB \: \: AD \: \: BD \: \: </p><p></p><p> \\ \\ \\ Similarly  \: ΔARQ  \: is \: similar \\ \: to \:  ΔADC \\ \\ </p><p> \frac{AQ}{AC} = \frac{AR}{AD} = \frac{RQ}{DC} - - - eq(ii) \\ \\ hence \: \\ \\ \\ \\ \\ \\ According \: to \: \\ equation \: (i) \: and \: (ii) </p><p> \\ \\ \\ </p><p>\frac{AR}{PR} = \frac{RQ}{AD} = \frac{BD}{DC} \\ \\ \\ \\ </p><p></p><p> but \\ \\  BD=DC (given) \\ \\ \\ </p><p></p><p>∴PR=RQ \\ \\ \\ </p><p></p><p>or  \\ \\ \\ AD bisects PQ at R </p><p></p><p></p><p>$

Answer 2

Answer:

=

sinxcosx

1

\rm \: = \: \dfrac{2}{2 \: sinx \: cosx} =

2sinxcosx

2

\rm \: = \: \dfrac{2}{sin \: 2x} =

sin2x

2

\rm \: = \: 2cosec2x = 2cosec2x