In fig.7.84, DEFG is a square and `angleBAC = 90°`. Prove that:

(i) `DeltaAGF~DeltaDBG` (ii) `DeltaAGF~ DeltaEFC `(iii) `DeltaDBG~ DeltaEFC` (iv) `DE^(2) = BD xx EC`

In the given figure DEFG is a square and angleBAC=90^(@) . Show that FG^(2) = BG xx FC .

In Fig. 4.128, D E F G is a square and /_B A C=90o . Prove that A G F D B G (ii) A G F E F C (FIGURE) (iii) D B G E F C (iv) D E^2=B DxxE C

In the given figure, ABCD is a quadrilateral in which AD = BC and angleDAB=angleCBA. Prove that (i) DeltaABD~=DeltaBAC, (ii) BD = AC, (iii) angleABD=angleBAC.

In the given figure, ABCD is a square and anglePQR=90^(@).If PB=QC=DR, prove that. (i) QB = RC (ii) PQ = QR (iii) angleQPR=45^(@)

In the given figure if angleB= 90^(@) and BD is perpendicular to AC then prove that: (i) triangleADB~triangleBDC (ii) triangleADB~triangleABC (iii) triangleBDC ~ triangleABC (iv) BD^(2)= AD xx DC (v) AB^(2)= AD xx AC (vi) BC^(2)= CD xx AC (vii) AB^(2)+BC^(2)= AC^(2)

ABCD is a quadrilateral in which and /_D A B\ =/_C B A (see Fig. 7.17). Prove that (i) DeltaA B D~=DeltaB A C (ii) B D\ =\ A C (iii) /_A B D\ =/_B A C

In the adjoining figure, ABCD is a square and PAB is an equilateral triangle. Find : (i) angleAPD (ii) anglePDC (iii) angleDPC (iv) Prove that DP = CP

In the adjoining figure D, E and F are the mid-points of the sides BC, CA and AB respectively of Delta ABC . Prove that: (i) square BDEF is a parallelogram (ii) area of Delta DEF = (1)/(4) xx " area of " Delta ABC (iii) area of square BDEF = (1)/(2) xx " area of " Delta ABC