Question

In the given figure,O is the centre of the circle,OD is perpendicular to AC, OE IS PERPENDICULAR TO BC AND OD IS EQUAL TO OE. SHOW THAT TRIANGLE DBA IS CONGRUENT TO TRIANGLE EAB.

Answer 1

ΔDBA  ≅ ΔEAB

Step-by-step explanation:

ΔOAD & ΔOBE

OA = OB  ( Radius)

OD = OE given

∠AOD = ∠BOE ( vertically opposite angle)

=> ΔOAD ≅ ΔOBE

=> AD = BE

AE =AO + OE

BD = BO + OD

AO = BO = Radius

OD = OE  given

=> AE = BD

in ΔDBA  & ΔEAB

AD = BE  Shown above

AE = BD  ( shown above)

∠ADB = ∠BEA = 90°

=> ΔDBA  ≅ ΔEAB

QED

Proved

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Answer 2

$\huge \underline \mathfrak{solution: - }$

Given,

OD=OE

Therefore, AC=BC(chords equidistant from centre are equal) ————(i)

In ΔACE and ΔBCD

AC=BC(from (i) )

$\angle \: c = \angle \: c(common \: angle)$

$CE=CD( \frac{1}{2} BC= \frac{1}{2} AC)$

$\triangle \: ACE \: ≅\triangle \: BCD$

AE =BD(CPCT)————(ii)

In ΔDBA and ΔEAB

AE=BD(from (ii) )

$DA=EB( \frac{1}{2}AC = \frac{1}{2} BC)$

AB=AB(common side)

$ΔDBA≅ΔEAB$

#hope it helps...

(For figure refer to the attachment)