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Triangles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsTriangles1 Mark
MCQ
In the adjoining figure, $\text{AB}\bot\text{BE}$ and $\text{FE}\bot\text{BE}.$ If $AB = FE$ and $BC = DE$, then.
  • A
    $\triangle\text{ABD}\cong\triangle\text{CEF}$
  • B
    $\triangle\text{ABD}\cong\triangle\text{ECF}$
  • ✓
    $\triangle\text{ABD}\cong\triangle\text{FEC}$
  • D
    $\triangle\text{ABD}\cong\triangle\text{EFC}$
✓

Answer

Correct option: C.
$\triangle\text{ABD}\cong\triangle\text{FEC}$
Given:
$AB = FE, BC = ED,$
$\text{AB}\bot\text{BE}$ and $\text{FE}\bot\text{BE}.$
To Prove: $AD = FC$
Proof: In $\triangle\text{ABD}$ and $\triangle\text{FEC},$
$AB = FE ...(1)$ (Given)
$\triangle\text{ABD}=\triangle\text{FEC} ...(2)$
Each$ = 90^{\circ}$
$BC = ED$ (Given)
$\Rightarrow BC + CD = ED + DC$
$\Rightarrow BD = EC ...(3)$
In view of $(1), (2)$ and $(3),$
$\triangle\text{ABD}\cong\triangle\text{FEC}$ using $SAS$ congruence rule.

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