Given In `Delta ABC ,angleB=90^(@)` and D is the mid-point of AC
Construction produce BD to E such that BD =DE and join EC
To Prove
Proof In `Delta ADB and DeltaCDE " " AD=DC " " [because "D is mid-point of" AC]`
BD =DE [by construction ]
and ` angle ADB = angleCDE` [vertically opposite angles ]
`therefore DeltaADB cong Delta CDE` [by SAS Congruence rule]
`rArr AB=EC` [by CPCT]
But `angleBAD and angle DCE ` are alternate angles
So EC|| AB and BC is a transversal
`therefore angleABC + angle BCE =180^(@)` [cointerior angles ]
`rArr 90^(@)+angle BCE = 180^(@)` [`because angleABC =90^(@)` given ]
`rArr angleBCE=180^(@)-90^(@)`
`rArr angleBCE=90^(@)`
In `Delta ABC and Delta ECB` AB=EC [proved above]
BC=CB [common side]
and `angleABC=angleECB ` [each `90^(@)`]
`therefore Delta ABC cong Delta ECB ` [by SAS congruence rule ]
`rArr` AC=EB [by CPCT ]
`rArr 1/2EB =1/2 AC` [dividing both sides by 2]
`rArr BD =1/2 AC `
