In the given figure , DEFG is a square and `angleBAC= 90^(@)`
prove that
(i) `triangleAGF~ triangleDBG`
`(ii) triangleAGF~triangleEFC`
(iii) `triangleDBG ~ triangleEFC`
(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 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)

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, 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, angle ACB=90^(@) and CD bot AB . Prove that CD^(2)=BD*AD

In the adjoining figure, BD = CE and angleADB=angleAEC=90^(@) , prove that (i) DeltaABD cong DeltaACE (ii) ABC is an isosceles triangle.

In the given figure,AD = AE and AD^(2)=BD xx EC prove that: triangles ABD and CAE are similar.

In the given figure, AB"||"DC and angleD=angleC . Prove that : (i) AD = BC (ii) AC = BD

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