Question

11. In the given figure, BF and CE are of equal lengths and are
perpendicular to AD. If AB = DE, then prove that Angle ABC = Angle DEF​

Answer 1

Given:

BF = CE

BF ⊥ AD

CE ⊥ AD

AB = DE

To Prove:

∠ABC = ∠DEF

Proof:

In ΔBAF and ΔEDC,

∠BFA = 90° = ∠ECD = 90°

BF = CE (Given)

AB = DE (Given)

∴ By using the RHS congruency criterion we can say that:

⇒ ΔBAF ≅ ΔEDC.

Therefore by using CPCT we get,

⇒ AF = CD.

We know that AF = CD.

⇒ AF = CD

Add FC on both sides.

⇒ AF + FC = CD + FC

⇒ AC = FD ➝ (2)

In ΔBFC and ΔECF,

BF = CE (Given)

∠BFC = ∠ECF = 90° (Given)

FC = FC (Common side)

∴ By using the SAS congruency criterion we can say that:

⇒ ΔBFC ≅ ΔECF.

Therefore by using CPCT we get,

⇒ BC = FE ➝ (1)

In ΔABC and ΔDEF,

AB = DE (Given)

AC = FD (Proved in 1)

BC = FE (Proved in 2)

∴ By using the SSS congruency criterion we can say that:

⇒ ΔABC ≅ ΔDEF.

Therefore by using CPCT (Corresponding parts of congruent triangles) we get,

⇒ ∠ABC = ∠DEF

Hence proved.