Grade 128 grade maths

see attachment and explain it

Shivam

11 Years agoGrade 12

1 Answer

Nishant Vora

Approved Tutor Answer11 Years ago

Hello Student, Please find the solution

Consider triangle FAC and EBC
$\angle FAC = \angle EBC = 90$
$\angle FCA = \angle ECB (vertically opposite angles)$
Hence by AA traingle FAC and EBC are similar
so $\frac{AF}{BE} = \frac{AC}{BC}........... (Eqn 1)$

Now consider Triangle OAF and OBD
$\angle FOA = \angle DOB$
$\angle OAF = \angle OBD = 90$
So again by AA similarity $\Delta OAF = \Delta OBD$

$\frac{OA}{OB} = \frac{AF}{BD} .........eqn 2$

From (1) and (2) and also BD=BE
Therefore, $\frac{AC}{BC} = \frac{OA}{OB}$

=> $\frac{OC-OA}{OB-OC} = \frac{OA}{OB}$

=> $\frac{OC-OA}{OA} = \frac{OB-OC}{OB}$

Therefore, $\frac{1}{OA} + \frac{1}{OB} = \frac{2}{OC}$