OA-OB-OC-OD- angle AOB-angle COD- Prove that AC-BD

Question

Toppr Admin · Accepted Answer

Given -x2220-AOB-x2220-COD-x2220-AOB-x2212-x2220-COB-x2220-COD-x2212-x2220-COB-x2234-x2220-AOC-x2220-DOBIn -x394-AOC and -x394-DOBi-OA-OB-Given-ii-OC-x2220-OD -Given-iii-x2220-AOC-x2220-DOB -Proved above-x2234-x394-AOC-x2245-x2220-DOB-SAS Axiom-x2234-AC-BD

$O A = O B, O C = O D$ , $∠ A O B = ∠ C O D$ . Prove that $A C = B D$

Given $∠ A O B = ∠ C O D ∠ A O B - ∠ C O B = ∠ C O D - ∠ C O B ∴ ∠ A O C = ∠ D O B$
In $Δ A O C$ and $Δ D O B$
$i) O A = O B$ (Given)
$i i) O C = ∠ O D$ (Given)
$i i i) ∠ A O C = ∠ D O B$ (Proved above)
$∴ Δ A O C ≅ ∠ D O B$
(SAS Axiom)
$∴ A C = B D$

OA=OB,OC=OD, ∠AOB=∠COD. Prove that AC=BD

Given ∠AOB=∠COD∠AOB−∠COB=∠COD−∠COB∴∠AOC=∠DOBIn ΔAOC and ΔDOBi)OA=OB(Given)ii)OC=∠OD (Given)iii)∠AOC=∠DOB (Proved above)∴ΔAOC≅∠DOB(SAS Axiom)∴AC=BD

$O A = O B, O C = O D$ , $∠ A O B = ∠ C O D$ . Prove that $A C = B D$

Given $∠ A O B = ∠ C O D ∠ A O B - ∠ C O B = ∠ C O D - ∠ C O B ∴ ∠ A O C = ∠ D O B$
In $Δ A O C$ and $Δ D O B$
$i) O A = O B$ (Given)
$i i) O C = ∠ O D$ (Given)
$i i i) ∠ A O C = ∠ D O B$ (Proved above)
$∴ Δ A O C ≅ ∠ D O B$
(SAS Axiom)
$∴ A C = B D$