Question

ABCD is a square.M is the midpoint on AB such that AM =MB.P and Q are points on sides AD and extended CB such that CM Perpendicular PQ .Show that area Of triangle CPM =area of triangle CQM.

Answer 1

Hi friend,...

in triangle PAM and triangle QBM
angle PMA = angle QMB ( vertically opposite angle )
AM=MB (M is the mid point of AB)
angle PAM = angle QBM (90 ° )
∆ PAM congruent to ∆ QBM ( SAS)
PM = QM ( CPCT)
in triangle CPM and CQM
PM=QM ( proved above)
angle CPM = angle CMQ(90° )
CM =CM ( common)
∆ CPM congruent ∆ CQM (SAS)
ar( ∆CPM) = par (∆ CQM) ( congruent ∆ s hâve equal areas.

hope this helps u....
:)