In DEFG is a square in a triangle ABC right angled at A.

Prove that

(i) ΔAGF ∼ Δ DBG (ii) Δ AGF ∼ Δ EFC

OR

In an obtuse Δ ABC (∠B is obtuse), AD is perpendicular to CB produced. Then prove that AC² = AB² + BC² + 2BC × BD.

Question

Philoid · Accepted Answer

(i) Given: DEFG is a square and ∠BAC = 90°

To Prove: DE² = BD × EC.

In AGF and DBG

∠GAF = ∠BDG [each 90°]

∠AGF = ∠DBG

[corresponding angles because GF|| BC and AB is the transversal]

∴AFG ~ DBG [by AA Similarity Criterion] …(1)

(ii) In AGF and EFC

∠GAF = ∠CEF [each 90°]
∠AFG = ∠ECF

[corresponding angles because GF|| BC and AC is the transversal]

∴AGF ~EFC [by AA Similarity Criterion] …(2)

OR

Given: AD ⊥ CB (produced)

To prove: AC² = AB² + BC² + 2BC • BD

In ∆ ADC, DC = DB + BC ……….(i)

First, in ∆ ADB,

Using Pythagoras theorem, we have

AB² = AD² + DB²⇒ AD² = AB² – DB² ……….(ii)

Now, applying Pythagoras theorem in ∆ ADC, we have

AC² = AD² + DC²

= (AB² – DB²) + DC² [Using (ii)]

= AB² – DB² + (DB + BC)² [Using (i)]

Now, ∵ (a + b)² = a² + b² + 2ab

∴ AC² = AB² – DB² + DB² + BC² + 2DB • BC

⇒ AC² = AB² + BC² + 2BC • BD

In DEFG is a square in a triangle ABC right angled at A.Prove that(i) ΔAGF ∼ Δ DBG (ii) Δ AGF ∼ Δ EFCORIn an obtuse Δ ABC (∠B is obtuse), AD is perpendicular to CB produced. Then prove that AC2 = AB2 + BC2 + 2BC × BD.